You CAN Ace Calculus

Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |

Arc Length (Vector Functions) |

Arc Length Function |

Arc Length Parameter |

Conservative Vector Fields |

Cross Product |

Curl |

Curvature |

Cylindrical Coordinates |

Lagrange Multipliers |

Line Integrals |

Partial Derivatives |

Partial Integrals |

Path Integrals |

Potential Functions |

Principal Unit Normal Vector |

Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

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Help Keep 17Calculus Free |
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Limits discussed on this page are approaching a constant, i.e. a specific finite number. In limit notation, they look like \(\displaystyle{ \lim_{x \rightarrow c}{~f(x)} }\) where \(c\) is a finite number. | ||||||||||||||||||

If you want to study the case where \(c\) is infinite, you need to go to the Limits At Infinity (Infinite Limits) page. (The panel below explains the terminology and how 17calculus defines finite and infinite limits.) | ||||||||||||||||||

## Finite Limits, Infinite Limits, Limits At Infinity . . . Terminology Explained
The use of the terms finite limits, infinite limits and limits at infinity are used differently in various books and your instructor may have their own idea of what they mean. In this panel, we will try to break down the cases and explain the various ways these terms can be used as well as how we use them here at 17calculus.
You can see where the confusion lies. The terms | ||||||||||||||||||

Studying limits that approach a constant is the best place to start to get an understanding of how limits work. Make sure you understand this Limit Key before going on. | ||||||||||||||||||

## Limit KeyOne key that you need to remember about limits is when you use
the limit notation \[ \lim_{x \rightarrow c}{ ~f(x) } \]
this means that Here is an explanation of how this concept is written mathematically. Look more carefully at the definition of the limit at the top of the page. Notice that it requires \( \delta \gt 0 \). This means that \( |x-c| > 0 \) and, therefore, x can never equal c. I know this seems like a minor point, but it isn't. If you remember this, you will have a good start on your way to understanding limits. |

General Steps To Evaluate Limits |
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*Step 1 [ substitution ]* - - To evaluate a limit at a constant, the first thing you will always want to try is direct substitution. From the Limit Key discussion, you know that a limit doesn't necessarily mean the value AT the point. However, if the function is continuous at the point in question, the limit will be equal to the function at that point. Plugging the number into the function will help you determine if that is the case.

If you plug in the number into the function and you get a finite number without a zero in the denominator, then you are done and that number is your answer.

However, if you get a zero in the denominator, there is more work to do. The next step is determined by what you have in the numerator. If you have a number other than zero in the numerator, then go to Step 2A, otherwise go to Step 2B.

*Step 2A ( zero denominator, non-zero numerator )* - - In this case get a non-zero number in the numerator and zero in the denominator. This case means that the limit is either \(+\infty\) or \(-\infty\). In this case you need to look at the behavior of the denominator very near zero to see if it stays negative or positive on both sides of the limit.

*Step 2B ( zero denominator AND zero numerator )* - - What's going here is that \(0/0\) is what we call indeterminate. Basically, that means it can't be determined, i.e. \( 0/0 \) can mean anything. It may mean \(0\), it may mean \( 27 \) or \( 19075 \) or infinity. You can't tell. So what we need to do is to work some algebra on the problem to get it in a different form. Usually this means factoring the numerator and denominator to get a common factor that will cancel. [ See the practice problems for demonstrations of this technique. ] Another option may be to use trig identities so that factoring becomes possible.

*Step 3* - - Once you have altered the form of the limit, use substitution again to see if anything has changed. If not, then try again by going back to step 2 ( whichever applies ) and try again.

From the above list of steps, we extrapolate four main techniques that you might use, depending on the situation. Here, we discuss each technique in detail.

Technique 1 - Substitution |
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We know from the concept of continuity, that if a function, \( f(x) \), is continuous at the point \( x=c \), then \(\displaystyle{ \lim_{x \to c}{[f(x)]} = f(c) }\). So this tells us, that if the function is continuous at the point at which we are taking the limit, then we can just substitute that point into the function to determine the limit and we are done.

We can also substitute the limit value into the function to help us determine what to do next, even if the function is not continuous at the point. Here is what substitution tells us.

**Case 1:** After substitution, if we get an indeterminate form, we need to use one (or more) of the techniques discussed below (factoring, rationalizing, using trig identities or, if you know derivatives, L'Hôpital's Rule).

**Case 2:** After substitution, if we get a fraction like \( c/0 \) where \( c \) is any **non-zero** finite number this indicates a possible asymptote, so we need to look at what is happening on either side of \( x=c \). There are 4 possibilities. (These graphs look at \(c=0\). However, the discussion holds regardless of the value of \(c\).)

\(\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }\) and \(\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }\) |
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The limit exists and \(\displaystyle{ \lim_{x \to c}{[f(x)]} = +\infty }\). |

\(\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }\) and \(\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }\) |
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The limit exists and \(\displaystyle{ \lim_{x \to c}{[f(x)]} = -\infty }\). |

\(\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }\) and \(\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }\) |
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The limit does not exist. |

\(\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }\) and \(\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }\) |
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The limit does not exist. |

**Note** - On this site (and in many textbooks), we say that when \(\displaystyle{ \lim_{x \to c}{[f(x)]} = \infty }\), the limit exists and the limit is infinity (similarly for negative infinity). Not everyone agrees with this and your textbook and/or instructor may see this differently. So ask your instructor for clarification.

Now let's use this technique on some practice problems.

\(\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 3}\left[\frac{5x^2-8x-13}{x^2-5}\right]=2}\) |

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}\)

Solution |
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You always want to try direct substitution first.

\(\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{5x^2-8x-13}{x^2-5} \right] }\) |

\(\displaystyle{\frac{5(9)-8(3)-13}{9-5} }\) |

\(\displaystyle{ \frac{45-24-13}{4} }\) |

\(\displaystyle{ \frac{8}{4} = 2 }\) |

In this case, the result is determinate. So the limit is 2.

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 3}\left[\frac{5x^2-8x-13}{x^2-5}\right]=2}\) |

close solution |

\(\displaystyle{\lim_{x\to1}{\frac{3x+5}{x+4}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to1}{\frac{3x+5}{x+4}}}\)

Solution |
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video by PatrickJMT

close solution |

\(\displaystyle{\lim_{x\to0}{(x^2+x-6)}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to0}{(x^2+x-6)}}\)

Solution |
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video by PatrickJMT

close solution |

\(\displaystyle{\lim_{x\to14}{(x+\sqrt{x+11})}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to14}{(x+\sqrt{x+11})}}\)

Solution |
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video by PatrickJMT

close solution |

\(\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }\)

Final Answer |
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\(\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} = 2 }\) |

Problem Statement |
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\(\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }\)

Solution |
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video by Krista King Math

Final Answer |
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\(\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} = 2 }\) |

close solution |

\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}=+\infty}\) |

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}\)

Solution |
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Direct substitution yields \(\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]} = \frac{-3}{0}}\). So the answer is \(+\infty\), \(-\infty\) or DNE (Does Not Exist). How do you know which one (without a graph)?

We need to determine the behavior of the function on either side of \(x=0\). On the left side of zero when \(x < 0\) and is very close to zero, the numerator is negative. We know this since *x* is very, very small, so \(x^4\) is positive but very small and \(5x\) is very small but negative. Both terms will be dominated by -3, so the numerator is negative. In the denominator, we need to look at the square root; \(x^2 \) is always positive and adding a positive number to 4 means that the number under the square root is always greater than 4, so the square root is always larger than 2. This makes the denominator negative. So we have a negative numerator and a negative denominator, so the limit \(\displaystyle{ \lim_{x \rightarrow 0^-}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }\)

We use similar logic when \(x > 0\), to get \(\displaystyle{ \lim_{x \rightarrow 0^+}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }\)

Since the limit from the right is equal to the limit from the left, the original limit is also \( +\infty \).

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}=+\infty}\) |

close solution |

Technique 2 - Factoring |
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Once we have substituted the limit value in the function and get the indeterminate form \( 0/0 \), we can try factoring and canceling to get a determinate form. This technique works best with polynomials or with terms that can factor.

When we get \( 0/0 \), we have a hole in the graph (a removable discontinuity). In order to understand why this works, we invoke the following theorem.

Theorem: Functions That Agree At All But One Point [proof] |
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Let \(c\) be a real number and let \( g(x) = h(x) \) for all \( x \neq c \) in an open interval containing \(c\). |

If \(\displaystyle{ \lim_{x \to c}{[g(x)]} }\) exists, then \(\displaystyle{ \lim_{x \to c}{[h(x)]} }\) also exists and \(\displaystyle{ \lim_{x \to c}{[h(x)]} = \lim_{x \to c}{[g(x)]} }\). |

What this theorem says is that if we have two functions that are identical, expect (possibly) at \(x=c\), and we know that one function has a limit that exists at \(c\), then the other one is guaranteed to have a limit that exists at \(c\) and that limit is the same as the limit of the first function.

This theorem is incredibly useful for functions with holes (our current topic). If we have a function with a hole, and if we can find a second function that is the same as first function, except at \( x=c\), that has a limit there, then we can use that second function to determine the limit of the first function. Usually, this second function is simpler than the first and the way we find it is by factoring and canceling common terms.

Here is a quick example.

Let \(\displaystyle{ f(x) = \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} }\) and
\( g(x) = x+1 \)

Notice that by canceling the \((x-1)\) term in \(f(x)\), we get \(g(x)\). However, \( f(x) \neq g(x) \) since the domains of the two functions are different. The domain of \(f(x)\) is \( x \neq 1 \) and the domain of \(g(x)\) is all real numbers.

However, we can say that \(\displaystyle{ \lim_{x \to 1}{[f(x)]} = \lim_{x \to 1}{[g(x)]} }\) by applying the above theorem. Notice the difference, the functions are not equal but the limits at \( x=1\) are equal. This is an important distinction.

Okay, let's try factoring on some practice problems.

Basic Problems |
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\(\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}=11/4}\) |

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}\)

Solution |
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First of all, you always want to try direct substitution. When you do this on this equation, you get \(0/0\) which is indeterminate.

There are at least two ways to solve this. Since this is a ratio of polynomials, we will use algebra. You can also use L'Hôpital's Rule.

First, we factor the numerator and denominator and cancel like terms. |

\(\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x^2-x-10}{x^2-4} \right] = \lim_{x \rightarrow 2} \left[ \frac{(x-2)(3x+5)}{(x-2)(x+2)} \right]}\) |

Now the \((x-2)\) term that appears in both the numerator and denominator can cancel. |

\(\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right]}\) |

Now let's try substitution again. |

\(\displaystyle{\lim_{x \rightarrow 2} \left[ \frac{3x+5}{x+2} \right] = \frac{3(2)+5}{2+2} = 11/4 }\) |

This technique will work only with polynomials that can be factored.

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}=11/4}\) |

close solution |

\(\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}\)

Final Answer |
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\(\displaystyle{ \lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] }= 108/7 }\) |

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}\)

Solution |
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First of all, you always want to try direct substitution. When you do this in this limit, you get \(0/0\) which is indeterminate.

There are at least two ways to solve this. We will work it using algebra (since this is a ratio of polynomials). (You can also use L'Hôpital's Rule, since direct substitution yields an indeterminate form.)

First, we factor the numerator and denominator and cancel like terms. |

\(\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{x^4-81}{2x^2-5x-3} \right] }\) |

\(\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x^2-9)(x^2+9)}{(x-3)(2x+1)} \right] }\) |

\(\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} \right]}\) |

Now the \((x-3)\) term that appears in both the numerator and denominator can cancel. |

\(\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] }\) |

Now let's try substitution again. |

\(\displaystyle{ \lim_{x \rightarrow 3} \left[ \frac{(x+3)(x^2+9)}{(2x+1)} \right] = \frac{(3+3)(3^2+9)}{(2(3)+1)} = 108/7}\) |

This technique will work only with polynomials and only if you can factor.

Final Answer |
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\(\displaystyle{ \lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] }= 108/7 }\) |

close solution |

\(\displaystyle{\lim_{x\to2}{\frac{x^2-5x+6}{x^2-4}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to2}{\frac{x^2-5x+6}{x^2-4}}}\)

Solution |
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video by Krista King Math

close solution |

\(\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}}}\)

Final Answer |
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\(\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}}=-3/2}\) |

Problem Statement |
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\(\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}}}\)

Solution |
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She made a mistake in notation here. In the last line of the first column, she wrote
\(\displaystyle{ \lim_{x \to 1}{\frac{x+2}{x-3}}}\) which is correct. However, in the first line of the next column, she should have written \(\displaystyle{\frac{1+2}{1-3}=\frac{3}{-2}}\) WITHOUT the limit as *x* goes to *1* in front of it. Once you substitute the value, you do not write the limit.

video by Krista King Math

Final Answer |
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\(\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x^2-4x+3}}=-3/2}\) |

close solution |

\(\displaystyle{\lim_{x\to-1}{\frac{x+1}{x^2-x-2}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to-1}{\frac{x+1}{x^2-x-2}}}\)

Solution |
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video by Krista King Math

close solution |

\(\displaystyle{\lim_{t\to-3}{\frac{t^2+6t+9}{t^2-9}}}\)

Problem Statement |
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\(\displaystyle{\lim_{t\to-3}{\frac{t^2+6t+9}{t^2-9}}}\)

Solution |
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video by Krista King Math

close solution |

\(\displaystyle{\lim_{z\to-2}{\frac{(z+2)^2}{z^4-16}}}\)

Problem Statement |
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\(\displaystyle{\lim_{z\to-2}{\frac{(z+2)^2}{z^4-16}}}\)

Solution |
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video by Krista King Math

close solution |

\(\displaystyle{\lim_{x\to2}{\frac{x^2-4}{x-2}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to2}{\frac{x^2-4}{x-2}}}\)

Solution |
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video by PatrickJMT

close solution |

\(\displaystyle{\lim_{h\to0}{\frac{(4+h)^2-16}{h}}}\)

Problem Statement |
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\(\displaystyle{\lim_{h\to0}{\frac{(4+h)^2-16}{h}}}\)

Solution |
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video by PatrickJMT

close solution |

\(\displaystyle{\lim_{x\to-1}{\frac{2x+2}{x+1}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to-1}{\frac{2x+2}{x+1}}}\)

Solution |
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video by Khan Academy

close solution |

\(\displaystyle{\lim_{x\to3}{\frac{x^2-6x+9}{x^2-9}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to3}{\frac{x^2-6x+9}{x^2-9}}}\)

Solution |
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video by Khan Academy

close solution |

\(\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x-1}}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\to1}{\frac{x^2+x-2}{x-1}}}\)

Solution |
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video by Khan Academy

close solution |

\(\displaystyle{\lim_{z\to2}{\frac{z^2+2z-8}{z^4-16}}}\)

Problem Statement |
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\(\displaystyle{\lim_{z\to2}{\frac{z^2+2z-8}{z^4-16}}}\)

Solution |
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video by Khan Academy

close solution |

Intermediate Problems |
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\(\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}=-1/48}\) |

Problem Statement |
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\(\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}}\)

Solution |
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The first thing you want to try here is to plug in -2 for *t*. However, in this limit, we get 0/0, which is an indeterminate form. So let's try factoring. But first we need to get the fraction in a different form.

In the numerator, we have \(\displaystyle{ \frac{1}{t} + \frac{1}{2} = \frac{2}{2t} + \frac{t}{2t} = \frac{t+2}{2t} }\).

So the limit looks like \(\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{ \frac{t+2}{2t}}{t^3+8} \right]} = \lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t^3+8)} \right]} }\).

Now we need to factor the denominator. Let's check quickly to make sure (t+2) is a factor. This is a great time to practice synthetic division. Using synthetic division, \(t^3+8\) factors into \((t+2)(t^2-2t+4)\).

So our limit is now \(\displaystyle{\lim_{t \rightarrow -2}{ \left[ \frac{t+2}{(2t)(t+2)(t^2-2t+4)} \right]} }\). We can cancel the (t+2) term in the numerator and denominator and get \(\displaystyle{\lim_{t \rightarrow -2} \left[ \frac{1}{(2t)(t^2-2t+4)} \right]}\).

Substituting -2 for *t* gives us \(\displaystyle{ \frac{1}{(-4)((-2)^2-2(-2)+4)} = \frac{1}{-4(4+4+4)} = -1/48 }\)

Final Answer |
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\(\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}=-1/48}\) |

close solution |

\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]}=-\infty}\) |

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]}}\)

Solution |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]} = \lim_{x\rightarrow 0}{\left[ \frac{x(x^2-7)}{x(x^2)} \right]} = \lim_{x \rightarrow 0}{\left[ \frac{x^2-7}{x^2} \right]} }\)

The trick here is determine if the limit is \( +\infty \), \( -\infty \) or DNE (does not exist). So we need to look at the limit as *x* approaches zero from the left and compare it to the limit from the right. If they are different, then the limit does not exist.

Let's look at the left side of zero first. When *x* is less than zero (negative), \(x^2\) is always positive. So the numerator is negative and the denominator is positive telling us that \(\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\).

Now, looking at the right side of zero, *x* is positive and \(x^2\) is also positive. Since the numerator is negative and the denominator is positive, we know that \(\displaystyle{ \lim_{x \rightarrow 0^+ }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\).

Since \(\displaystyle{ \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= \lim_{x \rightarrow 0^- }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\), the limit \(\displaystyle{ \lim_{x \rightarrow 0 }{\left[ \frac{x^2-7}{x^2} \right]}= -\infty }\)

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^3-7x}{x^3}\right]}=-\infty}\) |

close solution |

\(\displaystyle{\lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]}}\) does not exist. |

Problem Statement |
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\(\displaystyle{\lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]}}\)

Solution |
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Direct substitution yields \(0/0\) which is indeterminate.

Factoring yields

\(\displaystyle{ \lim_{z \rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]} = }\) \(\displaystyle{\lim_{z \rightarrow 1}{\left[\frac{(z-1)(z^2+z+1)}{(z-1)^2}\right]} = }\) \(\displaystyle{\lim_{z \rightarrow 1}{\left[\frac{z^2+z+1}{z-1}\right]}}\)

Direct substitution here gives us \(3/0\). So is the answer \(+\infty\) or \(-\infty\)? (without looking at a graph)

Well, looking at values close to 1, when \(z < 1\) the result is negative and when \(z>1\) the result is positive. So the answer is neither \(+\infty\) nor \(-\infty\) and the only correct is answer is that the limit does not exist.

Final Answer |
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\(\displaystyle{\lim_{z\rightarrow 1}{\left[\frac{z^3-1}{(z-1)^2}\right]}}\) does not exist. |

close solution |

\(\displaystyle{ \lim_{x\to1}{ \frac{x^3-1}{x^4-1} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x\to1}{ \frac{x^3-1}{x^4-1} } }\)

Solution |
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video by Krista King Math

close solution |

\(\displaystyle{\lim_{h\to0}{\frac{(3+h)^{-1}-3^{-1}}{h}}}\)

Problem Statement |
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\(\displaystyle{\lim_{h\to0}{\frac{(3+h)^{-1}-3^{-1}}{h}}}\)

Solution |
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video by PatrickJMT

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Advanced Problems |
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\(\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}=27}\) |

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}\)

Solution |
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Of course, the first thing we want to try is to plug in 27 directly for *x*. This gives us \(0/0\) which is indeterminate. So we need to do something else.

Upon initial inspection, this expression doesn't seem to be factorable. However, I am going to show you a trick. I don't know about you but I am not completely comfortable working with fractional powers. So I am going to substitute \(t\) for \(x^{1/3}\). We can do that as long we do the substitution for every expression containing *x*. So what do we get?

\(\displaystyle{ \frac{x-27}{x^{1/3}-3} = \frac{t^3-27}{t-3} }\)

Notice that since \(t=x^{1/3}\), we can cube both sides to get \(t^3=x\).

Using synthetic division, we can factor the numerator which becomes \((t^3-27)=(t-3)(t^2+3t+9)\).

Now the expression is \(\displaystyle{ \frac{(t-3)(t^2+3t+9)}{t-3}}\) and we can easily see that the \(t-3\) will cancel.

So let's go back to x's since our limit is in terms of *x*.

\(\displaystyle{ \lim_{x \rightarrow 27}{\frac{x-27}{x^{1/3}-3}} }\) |

\(\displaystyle{ \lim_{x \rightarrow 27}{\frac{(x^{1/3}-3)(x^{2/3}+3x^{1/3}+9)}{x^{1/3}-3}} }\) |

\(\displaystyle{ \lim_{x \rightarrow 27}{\frac{(x^{2/3}+3x^{1/3}+9)}{1}} }\) |

\( 27^{2/3}+3(27)^{1/3}+9 = 27 \) |

*Note* - - We could have used L'HÃ´pital's Rule here, which would have been much easier.

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}=27}\) |

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\(\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}=4/3}\) |

Problem Statement |
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\(\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}\)

Solution |
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Direct substitution yields \(0/0\), which is indeterminate. To solve this exercise we need to do a trick.

We want to see if we can factor this problem to cancel terms in the numerator and denominator. I don't know about you but I have a hard time working with fractional powers. So I am going to use substitution to remove fractional powers.

In this problem, I am going to substitute for *x*. I want to get rid of both the 1/3rd and 1/4th powers. So, to do that I am going to let \(t=x^{1/12}\). I chose 1/12 since 1/12=(1/3)(1/4). You will see in a minute why this works.

Okay, so we have \(t=x^{1/12}\) which means \(t^{12}=x\) (by taking the 12th power of both sides).

Now, let's substitute.

\(\displaystyle{ \frac{x^{1/3}-1}{x^{1/4}-1} = \frac{(t^{12})^{1/3}-1}{(t^{12})^{1/4}-1} = \frac{t^4-1}{t^3-1} }\)

Okay, so I can factor the numerator and denominator here (use synthetic division) and I get

\(\displaystyle{ \frac{(t-1)(t^3+t^2+t+1)}{(t-1)(t^2+t+1)}}\)

Even though this is not completely factored, we can see that we can cancel (t-1) in this fraction, so we can stop factoring here.

Okay, let's go back to limit notation and convert back to *x*'s.

\(\displaystyle{ \lim_{x \rightarrow 1}{\frac{(x^{1/12}-1)(x^{1/4}+x^{1/6}+x^{1/12}+1)}{(x^{1/12}-1)(x^{1/6}+x^{1/12}+1)}} }\) |

\(\displaystyle{\lim_{x \rightarrow 1}{\frac{x^{1/4}+x^{1/6}+x^{1/12}+1}{x^{1/6}+x^{1/12}+1}}}\) |

Now when we substitute 1 for *x*, we get \(\displaystyle{ \frac{1+1+1+1}{1+1+1} = 4/3 }\)

*Note* - - We could have used L'HÃ´pital's Rule here, which would have been much easier.

Final Answer |
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\(\displaystyle{\lim_{x\rightarrow 1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}=4/3}\) |

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Technique 3 - Rationalizing |
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This is one of the more interesting techniques and one that you are extremely likely to need to use on an exam. The idea is that you are given a limit that is indeterminate containing a square root term. You rationalize either the numerator or denominator (whichever location has the square root) to remove the square root. This will usually lead to a factor that can then be canceled so that the result is no longer indeterminate.

This technique depends on the idea demonstrated in this example: \( (\sqrt{x} - 1)(\sqrt{x}+1) = x-1\). Notice that the square root term is gone. In general, given a term that contains a square root, multiply by an identical term except for one sign change. For example, if you are given \( \sqrt{x+3} + 7 \), multiplying by \( \sqrt{x+3} - 7 \) will give you \( x+3 - 49 = x-46 \). It works the other way too, i.e. \( 3 - \sqrt{3x} \), multiplied by \( 3 + \sqrt{3x} \) yields \( 9 - 3x \).

Now, you can't just go around multiplying terms by whatever you want. In the problems where you will use this technique, you need to have the indeterminate form \(0/0\) and you multiply the numerator and denominator by the same term, essentially multiplying the entire function by \(1\). This is a valid operation that does not change the problem. Let's do a quick example because I want to show you something to watch out for.

\(\displaystyle{ \lim_{x \to 2}{\frac{\sqrt{x+2}-2}{x-2}} }\) |
First, let's carefully look at the fraction. Notice in the denominator we have a simple polynomial. We will leave that like it is for now. In the numerator we have a term with a square root. That is what we are looking for when using this technique. It doesn't matter if the term is in the numerator or denominator. We just want the term to have a square root. |

\(\displaystyle{ \lim_{x \to 2}{\frac{\sqrt{x+2}-2}{x-2}}\frac{\sqrt{x+2}+2}{\sqrt{x+2}+2} }\) |
In this step, we took the term with the square root and built a new term by changing the sign between the square root term and the other term. Then we multiplied the numerator and denominator by this new term. In this case, from \(\sqrt{x+2}-2\) we built \(\sqrt{x+2}+2\) |

So what have we accomplished here? It may seem like we haven't done anything since, essentially, we just multiplied by | |

\(\displaystyle{ \lim_{x \to 2}{\frac{x-2}{(x-2)(\sqrt{x+2}+2)}} }\) |
In this step, we multipled out the numerator and, this is what I wanted to warn you about, we left the denominator alone by leaving it factored and NOT multiplying it out. This is critical since the next step relies on this. We chose to multiply out the numerator, not because it is the numerator, but because it contains the term \(\sqrt{x+2}-2\) we used to build the new term \(\sqrt{x+2}+2\). |

\(\displaystyle{ \lim_{x \to 2}{\frac{1}{\sqrt{x+2}+2}} = 1/4 }\) | Now we cancel the common term \(x-2\) in the numerator and denominator, substitute and simplify to get our final answer. Notice that if we had multiplied out the denominator, we would not have been able to see what to cancel and that is the point of this technique, to cancel out the term that makes the numerator and denominator both zero, leaving us with our answer. |

**Comment** - -

This technique also works when you have two square roots in a term. For example, \( \sqrt{x+3} - \sqrt{x-5} \). The rationalization term will be \( \sqrt{x+3} + \sqrt{x-5} \).

Okay, try these practice problems.

\(\displaystyle{\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4}\right]}}\)

Problem Statement |
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\(\displaystyle{\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4}\right]}}\)

Final Answer |
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\(\displaystyle{\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4}\right]}=-1/6}\) |

Problem Statement |
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\(\displaystyle{\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4}\right]}}\)

Solution |
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As usual, you want to plug in 4 for *y* and see what you get. In this limit, you get 0/0, which is indeterminate. Our next idea might be to factor. However, there doesn't appear to be anything we can factor here. So, we have to do something called rationalizing.

To do this, we take the term with the square root in it and multiply the numerator and denominator by its conjugate. In this case, the conjugate of \(3-\sqrt{y+5}\) is \(3+\sqrt{y+5}\). The conjugate is obtained by changing the sign between the terms.

So the limit becomes

\(\displaystyle{\lim_{y \rightarrow 4}{\frac{3-\sqrt{y+5}}{y-4}}=\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4} \right] \left[ \frac{3+\sqrt{y+5}}{3+\sqrt{y+5}} \right] } }\)

Keep in mind that we need to multiply the numerator and denominator by the SAME EXPRESSION. In essence, this means we are multiplying by 1. Otherwise we change the problem (which we don't want to do).

Okay, so now what do we do? Now we multiply out the numerator and leave the denominator factored (you will see why in a minute).

The numerator is \(\displaystyle{ (3-\sqrt{y+5})(3+\sqrt{y+5}) = 9 - 3\sqrt{y+5} + 3\sqrt{y+5} - (y+5) }\). Do you see where all those terms came from?

Now, notice that there are two terms that look like \(3\sqrt{y+5}\) with different signs. So they cancel each other. So we are left with \(9-(y+5) = 9-y-5 = 4-y\). That's what the numerator reduces to. Our limit now looks like this.

\(\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{4-y}{(y-4)(3+\sqrt{y+5})} \right] } }\)

We are almost there. It looks like we have a term to cancel in the numerator and denominator but something is different. One of them is \(4-y\) and the other is \(y-4\). So they are slightly different. We need to do one more thing before we can cancel (the terms have to be EXACTLY the same to cancel). We need to get \(4-y\) to be \(y-4\). Well, what if we just switch the terms. If we do that we get \(-y+4\) and that is obviously not \(y-4\). (Remember, the sign in front of each term 'sticks' to that term, so when we switch the terms, we need to keep the negative sign with the *y*.)

To change \((-y+4)\) to \((y-4)\) we need to factor out a -1; (-y+4) = -1(y-4) giving us

\(\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{-1(y-4)}{(y-4)(3+\sqrt{y+5})} \right] } }\).

Now you can see why we didn't want to multiply out the denominator. If we had, it would not have been obvious that we can cancel the \((y-4)\) in numerator and denominator. Doing so we get

\(\displaystyle{ \lim_{y \rightarrow 4}{ \left[ \frac{-1}{(3+\sqrt{y+5})} \right] } }\) and substituting 4 for *y* gives us \(\displaystyle{ \frac{-1}{3+\sqrt{4+5}} = -1/6 }\).

Note:

This technique works whether the square root term is in the numerator or denominator and it doesn't matter if the square root term appears first or second. Just change one of the signs and not the other.

Final Answer |
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\(\displaystyle{\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4}\right]}=-1/6}\) |

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\(\displaystyle{ \lim_{x \to 0}{ \frac{\sqrt{x+1}-1}{x} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 0}{ \frac{\sqrt{x+1}-1}{x} } }\)

Solution |
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video by MIP4U

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\(\displaystyle{ \lim_{x \to 4}{ \frac{4-x}{\sqrt{x}-2} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 4}{ \frac{4-x}{\sqrt{x}-2} } }\)

Solution |
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video by MIP4U

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\(\displaystyle{ \lim_{x \to 9}{ \frac{\sqrt{x}-3}{x-9} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 9}{ \frac{\sqrt{x}-3}{x-9} } }\)

Solution |
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video by MIP4U

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\(\displaystyle{ \lim_{x \to 9}{ \frac{9-x}{3-\sqrt{x}} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 9}{ \frac{9-x}{3-\sqrt{x}} } }\)

Solution |
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video by ALEd

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\(\displaystyle{ \lim_{x \to 0}{ \frac{\sqrt{x+3}-\sqrt{3}}{x} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 0}{ \frac{\sqrt{x+3}-\sqrt{3}}{x} } }\)

Solution |
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video by ALEd

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\(\displaystyle{ \lim_{x \to 49}{\frac{49-x}{7-\sqrt{x}} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 49}{\frac{49-x}{7-\sqrt{x}} } }\)

Solution |
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video by ALEd

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\(\displaystyle{ \lim_{x \to 0}{ \frac{2-\sqrt{x+4}}{x} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 0}{ \frac{2-\sqrt{x+4}}{x} } }\)

Solution |
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video by ALEd

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\(\displaystyle{ \lim_{x \to 7}{\frac{\sqrt{x+2}-3}{x-7} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 7}{\frac{\sqrt{x+2}-3}{x-7} } }\)

Solution |
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video by PatrickJMT

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\(\displaystyle{ \lim_{x \to 0 }{ \frac{x}{\sqrt{x+9}-3} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{x \to 0 }{ \frac{x}{\sqrt{x+9}-3} } }\)

Solution |
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video by PatrickJMT

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\(\displaystyle{ \lim_{h \to 0}{\frac{\sqrt{4+h}-2}{h} } }\)

Problem Statement |
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\(\displaystyle{ \lim_{h \to 0}{\frac{\sqrt{4+h}-2}{h} } }\)

Solution |
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video by Krista King Math

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\(\displaystyle{ \lim_{x\to4}{\frac{\sqrt{x+5}-3}{x-4}} }\)

Problem Statement |
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\(\displaystyle{ \lim_{x\to4}{\frac{\sqrt{x+5}-3}{x-4}} }\)

Solution |
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video by rootmath

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Technique 4 - Using Trig Identities and Special Trig Limits |
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In addition to using the standard trig identities to convert trig functions into other, more useful, forms, we can use these two special limits to determine the limit involving trig functions when we have an indeterminate form.

\(\displaystyle{ \lim_{\theta \to 0}{\frac{\sin(\theta)}{\theta}} = 1 }\) and
\(\displaystyle{ \lim_{\theta \to 0}{\frac{1-\cos(\theta)}{\theta}} = 0 }\)

Some identities that you will need are |
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\( \cos^2(\theta) + \sin^2(\theta) = 1 \) |

\(\tan(\theta) = \sin(\theta) / cos(\theta) \) |

\( \sec(\theta) = 1/\cos(\theta) \) |

\( \csc(\theta) = 1/\sin(\theta) \) |

Detailed discussion and practice problems using this technique can be found on the trig limits page.