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You CAN Ace Calculus

17calculus > limits > finite limits

### Calculus Main Topics

Limits

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

Finite Limits - Limits Approaching A Finite Value

### 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.

When we talk about limits, we are looking at the $$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$. The various terms apply to the description of $$c$$ and $$L$$ and are shown in the table below. The confusion lies with the terms finite limits and infinite limits. They can mean two different things.

$$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$

when

term(s) used

$$c$$ is finite

limits approaching a finite value or finite limits

$$c$$ is infinite $$\pm \infty$$

limits at infinity or infinite limits

$$L$$ is finite

finite limits

$$L$$ is infinite $$\pm \infty$$

infinite limits

You can see where the confusion lies. The terms finite limits and infinite limits are used to mean two different things, referring to either $$c$$ or $$L$$. It is possible to have $$c = \infty$$ and $$L$$ be finite. So is this an infinite limit or a finite limit? It depends if you are talking about $$c$$ or $$L$$.

How 17calculus Uses These Terms
The pages on this site are constructed based on what $$c$$ is, i.e. we use the terms finite limits and infinite limits based on the value of $$c$$ only ( using the first two rows of the table above and ignoring the last two ). This seems to be the best way since, when we are given a problem, we can't tell what $$L$$ is until we finish the problem, and therefore we are unable to determine what type of problem we have and know what techniques to use until we are done with the problem.

Important: Make sure and check with your instructor to see how they use these terms.

This Page - - 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 above explains the terminology and how 17calculus defines finite and infinite limits. )

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 Key

One key that you need to remember about limits is when you use the limit notation $\lim_{x \rightarrow c}{ ~f(x) }$ this means that x APPROACHES c but is never equal to c. This means x can get as close as it wants to but it will never actually equal c. That seems simple enough but it is extremely important to remember. There are many times when you can determine the limit by substituting x for c to calculate f(c). However, those are special cases that require special conditions and is not true in every case. There is probably a theorem in your textbook that tells the special conditions that must exist for you to be able to apply this.

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

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.

Individual Techniques

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

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 in the following panels (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.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$

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

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$

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

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$

The limit does not exist.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$

The limit does not exist.

12

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.

Practice 1

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

solution

Practice 2

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

solution

Practice 3

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

solution

Practice 4

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

solution

Practice 5

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

solution

Practice 6

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

solution

Technique 2 - Factoring

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 ]

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.

13

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

Practice 7

$$\displaystyle{\lim_{x\rightarrow 2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}$$

solution

Practice 8

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

solution

Practice 9

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

solution

Practice 10

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

solution

Practice 11

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

solution

Practice 12

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

solution

Practice 13

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

solution

Practice 14

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

solution

Practice 15

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

solution

Practice 16

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

solution

Practice 17

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

solution

Practice 18

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

solution

Practice 19

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

solution

 Intermediate Problems

Practice 20

$$\displaystyle{\lim_{t\rightarrow -2}{\left[\frac{\frac{1}{t}+\frac{1}{2}}{t^3+8}\right]}}$$

solution

Practice 21

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

solution

Practice 22

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

solution

Practice 23

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

solution

Practice 24

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

solution

Practice 25

$$\displaystyle{\lim_{x\rightarrow 27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}$$

solution

Practice 26

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

solution

Technique 3 - Rationalizing

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 1. True but we multipled by a special 1. See, what we are going to have in the numerator is $$(\sqrt{x+2}-2)(\sqrt{x+2}+2)$$. When we multiply that out, we get $$(\sqrt{x+2})(\sqrt{x+2}) + 2(\sqrt{x+2})-2(\sqrt{x+2})-4$$. Look at this closely. Notice that the two middle terms with the square roots cancel each other and that is our goal. Now, the first term becomes $$(\sqrt{x+2})(\sqrt{x+2}) = x+2$$ and when completely simplified, the numerator is now $$x+2-4=x-2$$. This technique is what we call rationalizing. $$\displaystyle{ \lim_{x \to 2}{\frac{x-2}{(x-2)(\sqrt{x+2}+2)}} }$$ 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.

14

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 $$\sqrt{x+3} + \sqrt{x-5}$$.

Practice 27

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

solution

Practice 28

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

solution

Practice 29

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

solution

Practice 30

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

solution

Practice 31

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

solution

Practice 32

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

solution

Practice 33

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

solution

Practice 34

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

solution

Practice 35

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

solution

Practice 36

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

solution

Practice 37

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

solution

Practice 38

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

solution

Technique 4 - Using Trig Identities and Special Trig Limits

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

$$\cos^2(\theta) + \sin^2(\theta) = 1$$

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

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

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

Detail discussion and practice problems can be found on the trig limits page.