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17Calculus Limits - Solve By Rationalizing

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Evaluating Limits By Rationalizing

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 is the third technique for evaluating finite limits. It is very powerful and you will use it a lot in conjunction with substitution.

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. Let's do a quick demonstration.

Let's multiply \( ( \sqrt{x} - 1 ) \) by \( ( \sqrt{x} + 1 ) \). Notice that the terms are the same except for the sign in the middle.

\( ( \sqrt{x} - 1 )( \sqrt{x} + 1 ) \)

\( (\sqrt{x})(\sqrt{x}) - 1(\sqrt{x}) + 1(\sqrt{x}) -1(1) \)

The two middle terms cancel out (which is the point of this technique).

\( x - 1 \)

Bottom Line: \( ( \sqrt{x} - 1 )( \sqrt{x} + 1 ) = x-1 \)

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.

Example

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

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.

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 this technique on these practice problems.

Practice

Use rationalizing to solve these limits, giving your answers in exact, simplified form.

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4}\right]}}\)

Solution

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
\(\displaystyle{\lim_{y \rightarrow 4}{\frac{3-\sqrt{y+5}}{y-4}} = }\) \(\displaystyle{ \lim_{y\rightarrow 4}{\left[\frac{3-\sqrt{y+5}}{y-4} \right] \left[ \frac{3+\sqrt{y+5}}{3+\sqrt{y+5}} \right] } }\)
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 \( (3-\sqrt{y+5})(3+\sqrt{y+5}) = \) \( 9 - 3\sqrt{y+5} + 3\sqrt{y+5} - (y+5) \). 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 like this \((-y+4) = -1(y-4)\) 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

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

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{ \lim_{x \to 0}{ \frac{\sqrt{x+1}-1}{x} } }\)

Solution

MIP4U - 1829 video solution

video by MIP4U

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{ \lim_{x \to 4}{ \frac{4-x}{\sqrt{x}-2} } }\)

Solution

MIP4U - 1830 video solution

video by MIP4U

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x\to 9}{\frac{\sqrt{x}-3}{x-9}}}\)

Solution

MIP4U - 1831 video solution

video by MIP4U

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x\to 9}{\frac{9-x}{3-\sqrt{x}}}}\)

Solution

ALEd - 1832 video solution

video by ALEd

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x\to 0}{\frac{\sqrt{x+3}-\sqrt{3}}{x}}}\)

Solution

ALEd - 1833 video solution

video by ALEd

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x\to 49}{\frac{49-x}{7-\sqrt{x}}}}\)

Solution

ALEd - 1834 video solution

video by ALEd

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x\to 0}{\frac{2-\sqrt{x+4}}{x}}}\)

Solution

ALEd - 1835 video solution

video by ALEd

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x \to 7}{\frac{\sqrt{x+2}-3}{x-7}}}\)

Solution

PatrickJMT - 1836 video solution

video by PatrickJMT

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x\to 0 }{\frac{x}{\sqrt{x+9}-3}}}\)

Solution

PatrickJMT - 1837 video solution

video by PatrickJMT

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{h\to 0}{\frac{\sqrt{4+h}-2}{h}}}\)

Solution

Krista King Math - 1838 video solution

video by Krista King Math

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

Problem Statement

Evaluate this limit by rationalizing. \(\displaystyle{\lim_{x\to4}{\frac{\sqrt{x+5}-3}{x-4}}}\)

Solution

rootmath - 1839 video solution

video by rootmath

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Practice Instructions

Use rationalizing to solve these limits, giving your answers in exact, simplified form.

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