(L8) L’Hospital’s Rule#

By the end of the lesson you will be able to:

  • identify limits in indeterminate form

  • apply l’Hospital’s Rule to evaluate them (where applicable)

Lecture Videos#

Not-Tested Topic Videos#

These are videos for more advanced topics on Indeterminate Forms that you will not be tested on. Although you still might helpful and interesting!

Indeterminate Forms#

So far we have primarily looked at indeterminate quotients in much of our work on limits. Let’s formalize these definitions a little bit.

We say the limit \(\displaystyle \lim_{x\to a} \dfrac{f(x)}{g(x)}\) is a:

\(0/0\) - Indeterminate Form

\(\dfrac{0}{0}\) Indeterminate Form provided both:

\[ \lim_{x\to a} f(x)=0 \qquad \text{and} \qquad \lim_{x\to a} g(x)=0 \]

\(\infty/\infty\) - Indeterminate Form

\(\dfrac{\infty}{\infty}\) Indeterminate Form provided both:

\[ \lim_{x\to a} f(x)=\pm \infty \qquad \text{and} \qquad \lim_{x\to a} g(x)=\pm \infty \]

Note: Similar definitions exist for limits where \(x\to a^+\), \(x\to a^-\), \(x\to \infty\), and \(x\to -\infty\). That is, we have similar definitions for right and left side limits as well as limits at infinity.

L’Hospital’s Rule#

l’Hospital’s Rule

If we have a \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\) indeterminate form then:

\[ \lim_{x\to a} \dfrac{f(x)}{g(x)} = \lim_{x\to a} \dfrac{f'(x)}{g'(x)} \]

provided this limit exists or is infinite.

Work to Show

If we use l’Hospital’s Rule, we need to do 2 things:

  1. First verifty that our limit is of the correct indeterminate form, so that the Rule can even be applied.

  2. Explicitly indicate in our work, where we actually apply the rule. We usually use an H above the equal sign like so:

\[ \lim_{x\to a} \dfrac{f(x)}{g(x)} \stackrel{\text{H}}{=} \lim_{x\to a} \dfrac{f'(x)}{g'(x)} \]

Note 1: Similar rules exist for right and left side limits as well as limits at infinity.

Note 2: Make sure you do not confuse l’Hospital’s Rule with the Quotient Rule

  • Quotient Rule is used to calculate the derivative of \(\tfrac{f(x)}{g(x)}\)

  • l’Hospital’s Rule is used to calculate the limit of \(\tfrac{f(x)}{g(x)}\)

Indeterminate Quotients#

Example 1#

Evaluate the limit:

\[ \lim_{x\to 1} \dfrac{\ln x}{x-1} \]

Click through the tabs to see the steps of our solution.

Remember that whenever we evaluate a limit using l’Hospital’s Rule as part of our work we need to:

  1. Show that our limit is of the correct indeterminate form.

  2. Indicate where in our work we are using the Rule (usually with an H).

The first step is to identify the form of our limit. If we’re going to apply L’Hospital’s Rule, we first need to check that our limit is of the correct indeterminate form.

\[ \lim_{x\to 1} \dfrac{\ln x}{x-1} \longrightarrow \dfrac{\ln 1}{1-1} = \dfrac{0}{0} \]

We see that we have the indeterminate form \(\tfrac{0}{0}\). This is because the individual limits of both the numerator and denominator of the fraction are infinite:

\[ \lim_{x\to 1} \ln x = 0 \quad \text{and} \quad \lim_{x\to 1} x-1 = 0 \]

Note that this indeterminate form is not our answer but it does tell us we can use L’Hospital’s Rule!

Now that we have verified this is an appropriate indeterminate form, we can apply L’Hospital’s Rule. Indicate where you are using the rule in your work (we commonly use H above the equal sign).

\[ \lim_{x\to 1} \dfrac{\ln x}{x-1} \stackrel{\text{H}}{=} \lim_{x\to 1}\dfrac{ \big[\ln x\big]'}{\big[x-1\big]'} = \lim_{x\to 1}\dfrac{ \quad \tfrac{1}{x} \quad}{1} \]

Once we apply the rule, differentiate, and simplify, we have a brand new limit that needs to be investigated. So the process starts again:

  • Identify the form of this new limit.

  • Use L’Hospital’s Rule if applicable.

Investigating this new limit, we see that it is one we can actually do:

\[ \lim_{x\to 1}\dfrac{ \quad \tfrac{1}{x} \quad}{1} = \underbrace{\lim_{x\to 1}\dfrac{ 1}{x}}_{\text{continuous at } x=1}=\dfrac{1}{1} \]

And there’s our answer:

\[ \lim_{x\to 1} \dfrac{\ln x}{x-1} =1 \]

Example 2#

Evaluate the limit:

\[ \lim_{x\to \pi^-} \dfrac{\sin x}{1-\cos x} \]

Click through the tabs to see the steps of our solution.

It is important to first try and identify if the limit is an indeterminate form before jumping right into l’Hospital’s Rule.

The first step is to identify the form of our limit. If we’re going to apply L’Hospital’s Rule, we first need to check that our limit is of the correct indeterminate form.

\[ \lim_{x\to \pi^-} \dfrac{\sin x}{1-\cos x} \longrightarrow \dfrac{\sin \pi}{1-\cos \pi}= \dfrac{0}{1-(-1)} = \dfrac{0}{2} \]

We see that this is not an indeterminate form, so l’Hospital’s Rule cannot be applied.

If l’Hospital’s Rule cannot be used, then how to do we find the limit?

Well, we already have found it!

\[ \underbrace{\lim_{x\to \pi^-} \dfrac{\sin x}{1-\cos x}}_{\text{Continuous at }x=\pi} = \dfrac{\sin \pi}{1-\cos \pi}= \dfrac{0}{1-(-1)} = \dfrac{0}{2} = 0 \]

Remember \(\tfrac{0}{2}\) is a perfectly good number, the issue only happens when we try to divide by \(0\).

What are you still doing here? This example is done!

Example 3#

Evaluate the limit:

\[ \lim_{x\to\infty} \dfrac{e^x}{x^2} \]

Click through the tabs to see the steps of our solution.

It is important to first try and identify if the limit is an indeterminate form.

The first step is to identify the form of our limit. If we’re going to apply L’Hospital’s Rule, we first need to check that our limit is of the correct indeterminate form.

\[ \lim_{x\to\infty}\dfrac{ e^x}{x^2} \longrightarrow \dfrac{\infty}{\infty} \]

We see that we have the indeterminate form \(\tfrac{\infty}{\infty}\). This is because the individual limits of both the numerator and denominator of the fraction are infinite:

\[ \lim_{x\to\infty}e^x = \infty \quad \text{and} \quad \lim_{x\to\infty}x^2 = \infty \]

Note that this indeterminate form is not our answer but it does tell us we can use L’Hospital’s Rule!

Now that we have verified this is an appropriate indeterminate form, we can apply L’Hospital’s Rule. Indicate where you are using the rule in your work (we commonly use H above the equal sign).

\[ \lim_{x\to\infty}\dfrac{ e^x}{x^2} \stackrel{\text{H}}{=} \lim_{x\to\infty}\dfrac{ \big[e^x\big]'}{\big[x^2\big]'} = \lim_{x\to\infty}\dfrac{ e^x}{2x} \]

Once we apply the rule, differentiate, and simplify, we have a brand new limit that needs to be investigated. So the process starts again:

  • Identify the form of this new limit.

  • Use L’Hospital’s Rule if applicable.

Investigating this new limit, we see that it is also an indeterminate form:

\[ \lim_{x\to\infty}\dfrac{ e^x}{2x} \longrightarrow \dfrac{\infty}{\infty} \]

Applying L’Hospital’s Rule again gives us:

\[\begin{split} \begin{aligned} \lim_{x\to\infty}\dfrac{ e^x}{2x} & \stackrel{\text{H}}{=} \lim_{x\to\infty}\dfrac{ \big[e^x\big]'}{\big[2x\big]'} = \lim_{x\to\infty}\dfrac{ e^x}{2} \\ \end{aligned} \end{split}\]

Not done yet, we now have another brand new limit to calculate.

Finally, we have a limit that is not indeterminate:

\[ \lim_{x\to\infty}\dfrac{ e^x}{2} = \lim_{x\to\infty}\tfrac{1}{2} e^x = \infty \]

And there we have our answer! After applying L’Hospital’s Rule twice, we found the limit was infinite:

\[ \lim_{x\to\infty}\dfrac{ e^x}{x^2} = \infty \]

Example 4#

Evaluate the limit:

\[ \lim_{x\to\infty} \dfrac{\ln x}{\sqrt[3]{x}} \]

Click through the tabs to see the steps of our solution.

In this example, its important to remember that once we finish differentiating we should try to simplify the resulting fraction as much as possible first, before trying to evaluate the limit.

The first step is to identify the form of our limit. If we’re going to apply L’Hospital’s Rule, we first need to check that our limit is of the correct indeterminate form.

\[ \lim_{x\to\infty} \dfrac{\ln x}{\sqrt[3]{x}} \longrightarrow \dfrac{\infty}{\infty} \]

We see that we have the indeterminate form \(\tfrac{\infty}{\infty}\). This is because the individual limits of both the numerator and denominator of the fraction are infinite:

\[ \lim_{x\to\infty}\ln x = \infty \quad \text{and} \quad \lim_{x\to\infty}\sqrt[3]{x} = \infty \]

Note that this indeterminate form is not our answer but it does tell us we can use L’Hospital’s Rule!

Now that we have verified this is an appropriate indeterminate form, we can apply L’Hospital’s Rule. Indicate where you are using the rule in your work (we commonly use H above the equal sign).

\[ \lim_{x\to\infty}\dfrac{ \ln x}{\sqrt[3]{x}} \stackrel{\text{H}}{=} \lim_{x\to\infty}\dfrac{ \big[\ln x \big]'}{\big[\sqrt[3]{x}\big]'} = \lim_{x\to\infty}\dfrac{ \quad \tfrac{1}{x} \quad}{\tfrac{1}{3}x^{-2/3}} \]

Once we apply the rule, differentiate, and simplify, we have a brand new limit that needs to be investigated. So the process starts again:

  • Identify the form of this new limit.

  • Use L’Hospital’s Rule if applicable.

Once we finish differentiating, we should try to simplify our new limit as much as possible. We actually have a bunch of powers of \(x\) which probably can be consolidated in some way:

\[\begin{split} \lim_{x\to\infty}\dfrac{ \quad \tfrac{1}{x} \quad}{\tfrac{1}{3}x^{-2/3}}\cdot \dfrac{3x}{3x} &= \lim_{x\to\infty}\dfrac{ \quad \tfrac{3x}{x} \quad}{\tfrac{1}{3}x^{-2/3}\cdot (3x)} \\ & \\ &= \lim_{x\to\infty}\dfrac{ 3}{x^{1/3}} \\ \end{split}\]

Investigating this new limit, we see that it is not an indeterminate form:

\[ \lim_{x\to\infty}\dfrac{ 3}{x^{1/3}} \longrightarrow \dfrac{c}{\infty} \]

So L’Hospital’s Rule does not apply anymore, instead we can evaluate this limit using what we know about \(\tfrac{c}{\infty}\)-forms:

\[ \lim_{x\to\infty}\dfrac{ 3}{x^{1/3}} = 0 \]

Done!

Example 5#

Evaluate the limit:

\[ \lim_{x\to 0} \dfrac{\tan x -x}{x^3} \]

Click through the tabs to see the steps of our solution.

While we’re going through this example, don’t forget that when we’re differentiating, we may need to use some of our more advanced derivative rules like (product, quotient, and chain rule).

The first step is to identify the form of our limit. If we’re going to apply L’Hospital’s Rule, we first need to check that our limit is of the correct indeterminate form.

\[ \lim_{x\to 0} \dfrac{\tan x -x}{x^3} \longrightarrow \dfrac{\tan 0 - 0}{0^3} = \dfrac{0}{0} \]

We see that we have the indeterminate form \(\tfrac{0}{0}\). This is because the individual limits of both the numerator and denominator of the fraction are infinite:

\[ \lim_{x\to0}\tan x -x = 0 \quad \text{and} \quad \lim_{x\to 0 }x^3 = 0 \]

Note that this indeterminate form is not our answer but it does tell us we can use L’Hospital’s Rule!

Now that we have verified this is an appropriate indeterminate form, we can apply L’Hospital’s Rule. Indicate where you are using the rule in your work (with an H).

\[ \lim_{x\to 0} \dfrac{\tan x -x}{x^3} \stackrel{\text{H}}{=} \lim_{x\to 0}\dfrac{ \big[\tan x -x\big]'}{\big[x^3\big]'} = \lim_{x\to 0}\dfrac{ \sec^2 x -1 }{3x^2} \]

Once we apply the rule, differentiate, and simplify, we have a brand new limit that needs to be investigated. So the process starts again:

  • Identify the form of this new limit.

  • Use L’Hospital’s Rule if applicable.

Investigating this new limit, we see that it is also an indeterminate form:

\[ \lim_{x\to 0}\dfrac{ \sec^2 x -1 }{3x^2} \longrightarrow \dfrac{\sec^2 0 -1 }{3(0)^2} = \dfrac{1-1}{0}=\dfrac{0}{0} \]

Applying L’Hospital’s Rule again gives us:

\[\begin{split} \begin{aligned} \lim_{x\to 0}\dfrac{ \sec^2 x -1 }{3x^2} & \stackrel{\text{H}}{=} \lim_{x\to 0}\dfrac{ \big[\sec^2 x -1 \big]'}{\big[ 3x^2 \big]'} \\ & = \lim_{x\to 0}\dfrac{ 2\sec x (\sec x \tan x)}{6x} \\ & = \lim_{x\to 0}\dfrac{ 2\sec^2 x \tan x}{6x} \\ \end{aligned} \end{split}\]

Not done yet, we now have another brand new limit to calculate.

Another brand new limit we need to check. And this is also an indeterminate form:

\[ \lim_{x\to 0}\dfrac{ 2\sec^2 x \tan x}{6x} \longrightarrow \dfrac{2 \sec^2 0 \tan 0 }{6(0)} = \dfrac{2(1)^2(0)}{0}=\dfrac{0}{0} \]

Applying L’Hospital’s Rule yet again gives us:

\[\begin{split} \begin{aligned} \lim_{x\to 0}\dfrac{ 2\sec^2 x \tan x}{6x} & \stackrel{\text{H}}{=} \lim_{x\to 0}\dfrac{ \big[2\sec^2 x \tan x \big]'}{\big[ 6x \big]'} \\ & \\ & = \lim_{x\to 0}\dfrac{ \big[2\sec^2 x \big]' \tan x + 2\sec^2 x \big[ \tan x \big]'}{6} \\ & \\ & = \lim_{x\to 0}\dfrac{ 4\sec x (\sec x\tan x) \tan x + 2\sec^2 x \sec^2 x}{6} \\ & \\ & = \lim_{x\to 0}\dfrac{ 4\sec^2 x \tan x+ 2\sec^4 x}{6} \\ \end{aligned} \end{split}\]

Investigating this new limit, we see that it is finally not an indeterminate form:

\[\begin{split} \begin{aligned} \lim_{x\to 0}\underbrace{\dfrac{ 4\sec^2 x \tan x+ 2\sec^4 x}{6}}_{\text{continuous at } x=0} & = \dfrac{ 4\sec^2 0 \tan 0+ 2\sec^4 0}{6} \\ & \\ & = \dfrac{4(1)^2(0)+ 2(1)^4}{6} \\ & \\ &= \dfrac{2}{6} \\ \end{aligned} \end{split}\]

Finally, after three uses of l’Hospital’s Rule we have our answer:

\[ \lim_{x\to 0} \dfrac{\tan x -x}{x^3} = \dfrac{1}{3} \]

Done!

Indeterminate Products#

We say the limit \(\displaystyle \lim_{x\to a} {f(x)}\cdot{g(x)}\) is a:

\(0\cdot \infty\) - Indeterminate Form

\(0\cdot \infty\) Indeterminate Form provided both:

\[ \lim_{x\to a} f(x)=0 \qquad \text{and} \qquad \lim_{x\to a} g(x)=\pm \infty \]

Note: As usual, we have similar definitions for right and left side limits as well as limits at infinity.

In many cases, our strategy for dealing with indeterminate products is:

  1. Rewrite the product as a fraction.

  2. Use l’Hospital’s Rule (if possible).

We can rewrite the product as a fraction by choosing the first function to be in the numerator:

\[ f(x)\cdot g(x) \xrightarrow{convert} \dfrac{\quad f(x) \quad}{\tfrac{1}{g(x)}} \]

Essentially, this converts a \(0\cdot \infty\) indeterminate form into a \(\tfrac{0}{0}\) indeterminate form, so that we can then use l’Hospital’s Rule.

Remember that since \(g(x)\to \infty\), we must have \(\tfrac{1}{g(x)}\to 0\). Since \(\tfrac{1}{g(x)}\) is a \(\tfrac{c}{\infty}\) form.

We can also rewrite the product as a fraction by choosing the second function to be in the numerator:

\[ f(x)\cdot g(x) \xrightarrow{convert} \dfrac{\quad g(x) \quad}{\tfrac{1}{f(x)}} \]

Essentially, this converts a \(0\cdot \infty\) indeterminate form into a \(\tfrac{\infty}{\infty}\) indeterminate form, so that we can then use l’Hospital’s Rule.

Remember that since \(f(x)\to 0\), we must have \(\tfrac{1}{f(x)}\to \pm \infty\). Since \(\tfrac{1}{f(x)}\) is a \(\tfrac{c}{0}\) form.

Example 6#

Evaluate the limit:

\[ \lim_{x\to 0^+} \sqrt{x} \ln x \]

Click through the tabs to see the steps of our solution.

In this example, we are going to rewrite the product as a fraction, so that l’Hospital’s Rule might be applicable.

As with any limit, we first need to check to see what form we have (if we get a number then that is most likely our answer).

We’re interested in calculating the limit:

\[ \lim_{x\to 0^+} \sqrt{x} \ln x \]

So we begin by looking at each individual piece:

\[ \lim_{x\to 0^+} \sqrt{x} = 0 \qquad \text{and} \qquad \lim_{x\to 0^+} \ln x = -\infty \]

Since these pieces are being multiplied together in the original function, it means we have a \(0\cdot \infty\) indeterminate product.

Since this limit is an indeterminate product of the form \(0\cdot \infty\), we need to try rewriting this as a fraction. As we’ve discussed we have two options for how to do this:

  1. The first function can go into the numerator of the fraction.

  2. The second function can go into the numerator.

Let’s try putting the \((\ln x)\) term into the numerator.

\[ \sqrt{x} \ln x = \dfrac{ \quad \ln x \quad }{\tfrac{1}{\sqrt{x}}} \]

Why is this helpful? It just looks like we made the expression more complicated. Well, let’s consider what is happening in this fraction:

\[ \lim_{x\to 0^+} \dfrac{ \quad \ln x \quad }{\tfrac{1}{\sqrt{x}}} \longrightarrow \dfrac{-\infty}{\infty} \]

Which is an indeterminate \(\tfrac{\infty}{\infty}\)-form, meaning we can now use l’Hospital’s Rule.

(Remember that \(\tfrac{1}{\sqrt{x}}\) is a \(\tfrac{c}{0}\) form which is either \(+\infty\) or \(-\infty\).)

Now that we have verified this is an appropriate indeterminate form, we can apply L’Hospital’s Rule (using an H to indicate where we use the rule).

\[ \lim_{x\to 0^+} \dfrac{ \quad \ln x \quad }{\tfrac{1}{\sqrt{x}}} \stackrel{\text{H}}{=} \lim_{x\to 0^+}\dfrac{ \quad \big[\ln x \big]'\quad}{\big[x^{-1/2} \big]'} = \lim_{x\to 0^+}\dfrac{ \quad \tfrac{1}{x} \quad}{-\tfrac{1}{2}x^{-3/2}} \]

Once we apply the rule, differentiate, and simplify, we have a brand new limit that needs to be investigated. So the process starts again:

  • Identify the form of this new limit.

  • Use L’Hospital’s Rule if applicable.

Once we finish differentiating, we should try to simplify our new limit as much as possible. We actually have a bunch of powers of \(x\) which probably can be consolidated in some way:

\[\begin{split} \lim_{x\to 0^+}\dfrac{ \quad \tfrac{1}{x} \quad}{-\tfrac{1}{2}x^{-3/2}}\cdot \dfrac{2x}{2x} &= \lim_{x\to 0^+}\dfrac{ \quad \tfrac{2x}{x} \quad}{-\tfrac{1}{2}x^{-3/2}\cdot (2x)} \\ & \\ &= \lim_{x\to 0^+}\dfrac{ 2}{-x^{-1/2}} \\ & \\ &= \lim_{x\to 0^+}-2x^{1/2} \\ \end{split}\]

Investigating this new limit, we see that it is not an indeterminate form:

\[ \underbrace{\lim_{x\to 0^+}-2x^{1/2}}_{\text{continuous at } x=0} = -2(0)^{1/2} =0 \]

And there’s our answer:

\[ \lim_{x\to 0^+} \sqrt{x} \ln x = 0 \]

Indeterminate Differences#

We say the limit \(\displaystyle \lim_{x\to a} \big({f(x)}-{g(x)}\big)\) is an:

\(\infty - \infty\) - Indeterminate Form

\(\infty - \infty\) Indeterminate Form provided both:

\[ \lim_{x\to a} f(x)=\infty \qquad \text{and} \qquad \lim_{x\to a} g(x)= \infty \]

Note: As usual, we have similar definitions for right and left side limits as well as limits at infinity.

In many cases, we have two strategies for dealing with indeterminate differences:

  1. Rewrite the difference as a product.

  2. Rewrite the difference as a quotient, so that l’Hospital’s Rule can be used (if possible).

We have actually seen these indeterminate differences before. One of the little examples that we looked at was:

\[ \lim_{x\to \infty} (x^2-x) \]

This is an \(\infty-\infty\) indeterminate form. What we did was to try and factor the polynomial into a product:

\[ \lim_{x\to \infty} (x^2-x) = \lim_{x\to \infty} x\cdot(x-1) = \infty \]

That product is not an indeterminate form, since both factors are getting really large positive, their overall product gets very large positive as well. And so the limit is infinite.

Example 7#

Evaluate the limit:

\[ \lim_{x\to 1^+} \left( \dfrac{1}{\ln x}- \dfrac{1}{x-1} \right) \]

Click through the tabs to see the steps of our solution.

In this example, we are going to rewrite the difference as a single quotient, so that l’Hospital’s Rule might be applicable.

As with any limit, we first need to check to see what form we have (if we get a number then that is most likely our answer).

We’re interested in calculating the limit:

\[ \lim_{x\to 1^+} \left( \dfrac{1}{\ln x}- \dfrac{1}{x-1} \right) \]

So we begin by looking at each individual piece:

\[ \lim_{x\to 1^+} \dfrac{1}{\ln x} = \infty \qquad \text{and} \qquad \lim_{x\to 1^+} \dfrac{1}{x-1} = \infty \]

where each of these is a \(\tfrac{c}{0}\)-form.

Since we are subtracting these two fractions, we therefore have an \(\infty - \infty\) indeterminate difference.

One of our usual strategies when working with fractions is to combine them into one fraction using their common denominator. Doing this we get:

\[\begin{split} \begin{aligned} \dfrac{1}{\ln x}- \dfrac{1}{x-1} & = \dfrac{x-1}{(\ln x)(x-1)}- \dfrac{\ln x}{(\ln x)(x-1)} \\ & \\ & = \dfrac{x-1-\ln x}{(\ln x)(x-1)}\\ \end{aligned} \end{split}\]

Why is this helpful? Well, let’s consider what is happening in this single fraction:

\[ \lim_{x\to 1^+} \dfrac{x-1-\ln x}{(\ln x)(x-1)} \longrightarrow \dfrac{1-1-\ln 1}{(\ln 1)(1-1)} = \dfrac{0}{0} \]

Which is an indeterminate \(\tfrac{0}{0}\)-form, meaning we can now use l’Hospital’s Rule.

Now that we have verified this is an appropriate indeterminate form, we can apply L’Hospital’s Rule (using an H to indicate where we use the rule).

\[\begin{split} \begin{aligned} \lim_{x\to 1^+} \dfrac{x-1-\ln x}{(\ln x)(x-1)} & \stackrel{\text{H}}{=} \lim_{x\to 1^+}\dfrac{ \quad \big[x-1-\ln x \big]'\quad}{\big[(\ln x)(x-1) \big]'} \\ & \\ & = \lim_{x\to 1^+}\dfrac{ 1- \tfrac{1}{x} }{\tfrac{1}{x}(x-1)+\ln x} \\ \end{aligned} \end{split}\]

Once we apply the rule, differentiate, and simplify, we have a brand new limit that needs to be investigated. So the process starts again:

  • Identify the form of this new limit.

  • Use L’Hospital’s Rule if applicable.

Once we finish differentiating, we should try to simplify our new limit as much as possible. Since we see fractions stacked on top of fractions, that’s a good indication that there are things we can simplify.

In this case let’s multiply by \(\tfrac{x}{x}\) so we can hopefully cancel some of those \(\tfrac{1}{x}\) terms:

\[ \lim_{x\to 1^+}\dfrac{ 1- \tfrac{1}{x} }{\tfrac{1}{x}(x-1)+\ln x} \cdot \dfrac{x}{x} = \lim_{x\to 1^+}\dfrac{ x-1}{x-1+x\ln x} \]

Investigating this new limit, we see that it is also an indeterminate form:

\[ \lim_{x\to 1^+}\dfrac{ x-1}{x-1+x\ln x} \longrightarrow \dfrac{1-1 }{1-1+(1)\ln 1} = \dfrac{0}{0} \]

Applying L’Hospital’s Rule again gives us:

\[\begin{split} \begin{aligned} \lim_{x\to 1^+}\dfrac{ x-1}{x-1+x\ln x} & \stackrel{\text{H}}{=} \lim_{x\to 1^+}\dfrac{ \big[x-1 \big]'}{\big[ x-1+x\ln x \big]'} \\ & \\ & = \lim_{x\to 1^+}\dfrac{ 1 }{1+ \ln x + x\tfrac{1}{x}} \\ & \\ & = \lim_{x\to 1^+}\dfrac{ 1 }{1+ \ln x + 1} \\ & \\ & = \lim_{x\to 1^+}\dfrac{ 1 }{2+ \ln x} \\ \end{aligned} \end{split}\]

Not done yet, we now have another brand new limit to calculate.

Investigating this new limit, we see that it is not an indeterminate form:

\[ \underbrace{\lim_{x\to 1^+}\dfrac{ 1 }{2+ \ln x}}_{\text{continuous at } x=1} = \dfrac{ 1 }{2+ \ln 1} = \dfrac{1}{2} \]

And there’s our answer:

\[ \lim_{x\to 1^+} \left( \dfrac{1}{\ln x}- \dfrac{1}{x-1} \right) = \dfrac{1}{2} \]

Indeterminate Powers#

We say the limit \(\displaystyle \lim_{x\to a} {f(x)}^{g(x)}\) is a:

\(0^0\) Indeterminate Form

\(0^0\) Indeterminate Form provided both:

\[ \lim_{x\to a} f(x)=0 \qquad \text{and} \qquad \lim_{x\to a} g(x)= 0 \]

\(\infty^0\) Indeterminate Form

\(\infty^0\) Indeterminate Form provided both:

\[ \lim_{x\to a} f(x)=\infty \qquad \text{and} \qquad \lim_{x\to a} g(x)= 0 \]

\(1^{\infty}\) Indeterminate Form

\(1^{\infty}\) Indeterminate Form provided both:

\[ \lim_{x\to a} f(x)=1 \qquad \text{and} \qquad \lim_{x\to a} g(x)= \infty \]

Note (1): As usual, we have similar definitions for right and left side limits as well as limits at infinity.

Note (2): Note that a \(0^{\infty}\) form is not indeterminate.

Example 8#

Evaluate the limit:

\[ \lim_{x\to 0^+} x^{\sqrt{x}} \]

Click through the tabs to see the steps of our solution.

The strategy we are going to use here, starts similar to our technique for logarithmic differentiation:

  1. Apply $\ln \big( \cdot \big) to both sides.

  2. Simplify the right side.

  3. Calculate the limit of \(\ln y\)

  4. Use \(y=e^{\ln y }\) to calculate the limit of \(y\).

As with any limit, we first need to check to see what form we have (if we get a number then that is most likely our answer).

We’re interested in calculating the limit:

\[ \lim_{x\to 0^+} x^{\sqrt{x}} \]

So we begin by looking at each individual piece:

\[ \lim_{x\to 0^+} x =0 \qquad \text{and} \qquad \lim_{x\to 0^+} \sqrt{x} =0 \]

Since the base and exponent terms are both going to \(0\), this means we have a \(0^0\) indeterminate form.

With these indeterminate powers, we start by applying the natural log to both sides of our function:

\[\begin{split} \begin{aligned} y & = x^{\sqrt{x} }\\ & \\ \ln \big( y \big) & = \ln \big( x^{\sqrt{x} } \big) \end{aligned} \end{split}\]

We next simplify the right side using our log properties:

\[ \ln y = \sqrt{x} \ln x \]

Next, we calculate the limit of \(\ln y\).

\[ \lim_{x\to 0^+} \ln y = \lim_{x\to 0^+} \sqrt{x} \ln x \]

This is a brand new limit that we need to calculate, meaning we would need to go through the usual steps by first identifying the indeterminate form (if it even is indeterminate).

However, we did this problem back in Example 6, so we’ll just use our result (and work) from there.

\[ \lim_{x\to 0^+} \ln y = \lim_{x\to 0^+} \sqrt{x} \ln x = 0 \]

Now, we’re ready to calculate our desired limit, with a little help from the property that \(y=e^{\ln y}\).

\[\begin{split} \lim_{x\to 0^+} x^{\sqrt{x}} & = \lim_{x\to 0^+} y \\ & \\ & = \lim_{x\to 0^+} e^{\ln y} \\ & \\ & = \underbrace{e^{\;\left(\displaystyle \lim_{x\to 0^+} \ln y\right) }}_{\text{continuous function}} \\ & \\ & = e^{0} \\ & \\ & = 1 \\ \end{split}\]

Note that since \(e^{(\cdot)}\) is a continuous function, we can bring the limit inside this outer function.

And we have our desired answer!

\[ \lim_{x\to 0^+} x^{\sqrt{x}} = 1 \]