L'Hôpital's Rule and Indeterminate Forms

L'Hôpital's Rule

While L'Hôpital's Rule here is written for two sided limits at a finite number $a$, it is still valid for one-sided limits and limits at $\pm\infty$.

So, L'Hôpital's Rule tells us that if we have an indeterminate form $0/0$ or $\pm\infty/\infty$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. There are some important things to note about L'Hôpital's Rule.

It is also possible to use L'Hôpital's Rule more than once in the same problem. If your original limit gives, say, $0/0$, you can use L'Hôpital's Rule. Now let's say this new limit also gives $0/0$. You can use L'Hôpital's Rule again to obtain a second new limit that should still equal the original.

In certain instances, as we'll see, L'Hôpital's Rule can be used (meaning we get $0/0$ or $\pm\infty/\infty$) but does not end up being helpful to determine the limit. In these instances, we can usually revert back to methods discussed in the Limits Chapter.

There is a lot to say and more things to be careful about with L'Hôpital's Rule, but we will bring those up in some examples.

Now try one on your own!

Indeterminate Products

Suppose we'd like to determine the following limit:

Knowing which piece to move to the denominator depends on the actual expressions you are working with. One tip for thinking about this is to look at the fraction you end up with in preparation for L'Hôpital's Rule. One version might has slightly easier derivatives of the numerator and denominator than the other version. In general, more practice will get you used to the idea. And if one version doesn't work or isn't helpful, just go back and try the other version.

Let's use this idea to go back and determine that first limit introduced in this part.

So we see we were able to take a limit in the form of an indeterminate product ($0(-\infty)$) and rewrite it so that it became a limit of an indeterminate form of $0/0$ or $\pm\infty/\infty$. In doing so, we were allowed to then apply L'Hôpital's Rule.

We will see that much of what we do in the rest of this section is to somehow rewrite a limit so that it is in the form of $0/0$ or $\pm\infty/\infty$ and then apply L'Hôpital's Rule. If you keep this idea in mind, there will not be much you need to memorize; only knowing our end goal is to get a limit of either $0/0$ or $\pm\infty/\infty$ form.

Let's look at another example.

When faced with a product $(0)(\pm\infty)$ we can turn it into a quotient that will allow us to use L'Hôpital's Rule. However, as we saw in the last example we need to be careful with how we do that on occasion. Sometimes we can use either quotient and in other cases only one will work.

Indeterminate Differences

We have seen indeterminates of the form $\infty-\infty$ before (or equivalently $-\infty+\infty$). In certain instances, we found that factoring out the term which goes to $\infty$ (or $-\infty$) fastest was a good method for determining limits.

But there are other situations we might find ourselves in that still yeild the $\infty-\infty$ form. Let's see some examples and how we might be able to apply our algebra knowledge (and perhaps L'Hôpital's Rule if we meet the conditions).

As we saw, limits of the form $\infty-\infty$ can come from a variety of different function types. What is important is to recognize that $\infty-\infty$ is an indeterminate form and that more work needs to be done in order to determine the limit.

Indeterminate Exponentials

Now we come to the last group of indeterminate forms:

Unlike the other problems, we can't easily turn an exponential expression into a fraction; the exponent is fairly stuck. However, we can use logarithms (even if they're not there) and their properties to our advantage. Recall:

Notice that the exponent $b$ is no longer an exponent. In fact, $b\ln(a)$ is a product, and we know how to deal with limits of products (even ones which give indeterminate forms). The downside here is that $b\ln(a)$ is equal to $\ln(y)$, not $y$ itself. So if we want to get back to $y$, we need to apply an exponential:

In particular, this means that

It can be fairly messy to do all the work here in the exponent. What we will sometimes do is look and see exactly what is happening to just the exponent in terms of the limit. However, it is incredibly important to remember to go back to the orginial limit. If we find that $L=\displaystyle\lim_{x\to a}g(x)\ln\left(f(x)\right)$, then $\displaystyle\lim_{x\to a}f(x)^{g(x)}=e^L$.

Hopefully this will be easier to see with an example.

See if you can work out a problem yourself.

Practice Problems

Determine the following limits.

1. $\displaystyle \mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}}$
Click for solution


It may be tempting to jump right in and use L'Hôpital's Rule, but we need to verify that L'Hôpital's Rule can be used; that is, the limit is either in the form of $0/0$ or $\pm\infty/\infty$. A quick check with direct substitution shows us that,

$${\mbox{as}}\,\,\,x \to 2\hspace{0.5in}\frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} \to \frac{0}{0}$$

and so this is a form that allows the use of L'Hôpital's Rule. So let's apply L'Hôpital's Rule.

$$\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - 14x + 10}}{{2x + 1}}$$

At this point all we need to do is try the new limit (with direct substitution) and see if it can be done.

$$\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - 14x + 10}}{{2x + 1}} = \frac{{ - 6}}{5}$$

So, the limit can be done, and we done with the problem! The limit is then,

$$\mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 7{x^2} + 10x}}{{{x^2} + x - 6}} = - \frac{6}{5}.$$


Note that we could have just used our techniques from the limits chapter to determine this particular limit. The numerator and denominator of the original expression can be factored and a common factor from the numerator and denominator can cancel. That process looks like:

$$\begin{align*} \lim_{x\to2}\frac{x^3-7x^2+10x}{x^2+x-6} &= \lim_{x\to2}\frac{x(x-5)(x-2)}{(x+3)(x-2)} \\ &= \lim_{x\to2}\frac{x(x-5)}{x+3} \\ &= \frac{2(2-5)}{2+3} \\ &= -\frac{6}{5}. \end{align*}$$

We will not redo every problem here; this is only a remind that we don't always have to use L'Hôpital's Rule even when it applies and is helpful. L'Hôpital's Rule is simply another tool we can use for determining certain limits.



2. $\displaystyle \mathop {\lim }\limits_{w \to \, - 4} \frac{{\sin \left( {\pi w} \right)}}{{{w^2} - 16}}$
Click for solution


Let's begin by direct substitution. A quick check shows us that,

$${\mbox{as}}\,\,\,w \to \, - 4\hspace{0.5in}\frac{{\sin \left( {\pi w} \right)}}{{{w^2} - 16}} \to \frac{0}{0}$$

and so this is a form that allows the use of L'Hôpital's Rule. Let's apply L'Hôpital's Rule.

$$\mathop {\lim }\limits_{w \to \, - 4} \frac{{\sin \left( {\pi w} \right)}}{{{w^2} - 16}} = \mathop {\lim }\limits_{w \to \, - 4} \frac{{\pi \cos \left( {\pi w} \right)}}{{2w}}$$

Now let's try direct substitution on this new limit.

$$\mathop {\lim }\limits_{w \to \, - 4} \frac{{\sin \left( {\pi w} \right)}}{{{w^2} - 16}} = \mathop {\lim }\limits_{w \to \, - 4} \frac{{\pi \cos \left( {\pi w} \right)}}{{2w}} = \frac{{\pi \cos \left( { - 4\pi } \right)}}{{ - 8}} = \frac{\pi }{{ - 8}}$$

So, the limit can be done, and we done with the problem! The limit is then,

$$\mathop {\lim }\limits_{w \to \, - 4} \frac{{\sin \left( {\pi w} \right)}}{{{w^2} - 16}} = - \frac{\pi }{8}.$$


3. $\displaystyle \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {3t} \right)}}{{{t^2}}}$
Click for solution


Let's begin by directly seeing what happens as $t\to\infty$. A quick check shows us that,

$${\mbox{as}}\,\,\,t \to \infty \hspace{0.5in}\frac{{\ln \left( {3t} \right)}}{{{t^2}}} \to \frac{\infty }{\infty }$$

and so this is a form that allows the use of L'Hôpital's Rule. Let's apply L'Hôpital's Rule.

$$\mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {3t} \right)}}{{{t^2}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{1}{t}}}{{2t}} = \mathop {\lim }\limits_{t \to \infty } \frac{1}{{2{t^2}}}$$

Don’t forget to simplify after taking the derivatives. This can often be the difference between being able to do the problem or not. Now let's investigation what happens with this new limit as $t\to\infty.

$$\mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {3t} \right)}}{{{t^2}}} = \mathop {\lim }\limits_{t \to \infty } \frac{1}{{2{t^2}}} = 0$$

So, the limit can be done, and we done with the problem! The limit is then,

$$\mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {3t} \right)}}{{{t^2}}} = 0.$$


4. $\displaystyle \mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}}$
Click for solution


Let's begin by direct substitution. A quick check shows us that,

$${\mbox{as}}\,\,\,z \to 0\hspace{0.5in}\frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} \to \frac{0}{0}$$

and so this is a form that allows the use of L'Hôpital's Rule. Let's apply L'Hôpital's Rule. Before actually using L'Hôpital's Rule it might be better if we multiply out the denominator to make the derivative (and later steps a little easier). Doing this gives,

$$\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = \mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^4} + 2{z^3} + {z^2}}}.$$

Now let's apply L'Hôpital's Rule.

$$\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = \mathop {\lim }\limits_{z \to 0} \frac{{2\cos \left( {2z} \right) + 14z - 2}}{{4{z^3} + 6{z^2} + 2z}}$$

Now let's try direct substitution on this new limit. Here, we can see that

$${\mbox{as}}\,\,\,z \to 0\hspace{0.5in}\frac{{2\cos \left( {2z} \right) + 14z - 2}}{{4{z^3} + 6{z^2} + 2z}} \to \frac{0}{0}.$$

So, using L'Hôpital's Rule doesn’t give us a limit that we can do. However, the new limit is one that can use L'Hôpital's Rule on so let’s do that.

$$\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = \mathop {\lim }\limits_{z \to 0} \frac{{ - 4\sin \left( {2z} \right) + 14}}{{12{z^2} + 12z + 2}} = \frac{{14}}{2}$$

Okay, the second application of L'Hôpital's Rule gives us a limit we can do and so the value of this limit is,

$$\mathop {\lim }\limits_{z \to 0} \frac{{\sin \left( {2z} \right) + 7{z^2} - 2z}}{{{z^2}{{\left( {z + 1} \right)}^2}}} = 7.$$


5. $\displaystyle \mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}}}{{{{\bf{e}}^{1 - \,x}}}}$
Click for solution


Let's begin by directly seeing what happens as $x\to-\infty$. A quick check shows us that,

$${\mbox{as}}\,\,\,x \to \, - \infty \hspace{0.5in}\frac{{{x^2}}}{{{{\bf{e}}^{1 - \,x}}}} \to \frac{\infty }{\infty }$$

and so this is a form that allows the use of L'Hôpital's Rule. Let's apply L'Hôpital's Rule.

$$\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}}}{{{{\bf{e}}^{1 - \,x}}}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{2x}}{{ - {{\bf{e}}^{1 - \,x}}}}$$

At this point let’s try the limit and see if it can be done. However, in this case, we can see that,

$${\mbox{as}}\,\,\,x \to \, - \infty \hspace{0.5in}\frac{{2x}}{{ - {{\bf{e}}^{1 - \,x}}}} \to \frac{{ - \infty }}{{ - \infty }}.$$

So, using L'Hôpital's Rule doesn’t give us a limit that we can do. However, the new limit is one that can use L'Hôpital's Rule on so let’s do that.

$$\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}}}{{{{\bf{e}}^{1 - \,x}}}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{{2x}}{{ - {{\bf{e}}^{1 - \,x}}}} = \mathop {\lim }\limits_{x \to \, - \infty } \frac{2}{{{{\bf{e}}^{1 - \,x}}}} = 0$$

The second application of L'Hôpital's Rule gives us a limit we can do and so the value of the limit is,

$$\mathop {\lim }\limits_{x \to \, - \infty } \frac{{{x^2}}}{{{{\bf{e}}^{1 - \,x}}}} = 0.$$


6. $\displaystyle \mathop {\lim }\limits_{z \to \infty } \frac{{{z^2} + {{\bf{e}}^{4\,z}}}}{{2z - {{\bf{e}}^{\,z}}}}$
Click for solution


Let's begin by directly seeing what happens as $z\to\infty$. A quick check shows us that,

$${\mbox{as}}\,\,\,z \to \infty \hspace{0.5in}\frac{{{z^2} + {{\bf{e}}^{4\,z}}}}{{2z - {{\bf{e}}^{\,z}}}} \to \frac{\infty }{{ - \infty }}$$

and so this is a form that allows the use of L'Hôpital's Rule. Let's apply L'Hôpital's Rule.

$$\mathop {\lim }\limits_{z \to \infty } \frac{{{z^2} + {{\bf{e}}^{4\,z}}}}{{2z - {{\bf{e}}^{\,z}}}} = \mathop {\lim }\limits_{z \to \infty } \frac{{2z + 4{{\bf{e}}^{4\,z}}}}{{2 - {{\bf{e}}^{\,z}}}}$$

At this point let’s try the limit and see if it can be done. However, in this case, we can see that,

$${\mbox{as}}\,\,\,z \to \infty \hspace{0.5in}\frac{{2z + 4{{\bf{e}}^{4\,z}}}}{{2 - {{\bf{e}}^{\,z}}}} \to \frac{\infty }{{ - \infty }}.$$

So, using L'Hôpital's Rule doesn’t give us a limit that we can do. However, the new limit is one that can use L'Hôpital's Rule on so let’s do that.

$$\mathop {\lim }\limits_{z \to \infty } \frac{{{z^2} + {{\bf{e}}^{4\,z}}}}{{2z - {{\bf{e}}^{\,z}}}} = \mathop {\lim }\limits_{z \to \infty } \frac{{2z + 4{{\bf{e}}^{4\,z}}}}{{2 - {{\bf{e}}^{\,z}}}} = \mathop {\lim }\limits_{z \to \infty } \frac{{2 + 16{{\bf{e}}^{4\,z}}}}{{ - {{\bf{e}}^{\,z}}}}$$

Now let’s try the limit and see if it can be done. However, again we can see that,

$${\mbox{as}}\,\,\,z \to \infty \hspace{0.5in}\frac{{2z + 4{{\bf{e}}^{4\,z}}}}{{2 - {{\bf{e}}^{\,z}}}} \to \frac{\infty }{{ - \infty }}.$$

We might be thinking that we'll be stuck in a loop forever, but notice that these limits we get after applying L'Hôpital's Rule are getting progressively simpler as expressions. Let's apply L'Hôpital's Rule again, since it is of a correct form to do so, and we'll be sure to do any simplifying of the expression if possible.

$$\mathop {\lim }\limits_{z \to \infty } \frac{{{z^2} + {{\bf{e}}^{4\,z}}}}{{2z - {{\bf{e}}^{\,z}}}} = \mathop {\lim }\limits_{z \to \infty } \frac{{2z + 4{{\bf{e}}^{4\,z}}}}{{2 - {{\bf{e}}^{\,z}}}} = \mathop {\lim }\limits_{z \to \infty } \frac{{2 + 16{{\bf{e}}^{4\,z}}}}{{ - {{\bf{e}}^{\,z}}}}=\lim_{z\to\infty}\frac{64e^{4z}}{-e^z}=\lim_{z\to\infty}-64e^{3z}$$

Now investigate what happens as $z\to\infty$ with this new limit.

$$\mathop {\lim }\limits_{z \to \infty } \frac{{{z^2} + {{\bf{e}}^{4\,z}}}}{{2z - {{\bf{e}}^{\,z}}}}=\lim_{z\to\infty}-64e^{3z}=-\infty$$

While $-\infty$ is not a number, and so this limit does not exist, we still will say,

$$\mathop {\lim }\limits_{z \to \infty } \frac{{{z^2} + {{\bf{e}}^{4\,z}}}}{{2z - {{\bf{e}}^{\,z}}}}=-\infty.$$


7. $\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \displaystyle \frac{3}{t}} \right)} \right]$
Click for solution


Let's begin by directly seeing what happens as $t\to\infty$. A quick check shows us that,

$${\mbox{as}}\,\,\,t \to \infty \hspace{0.5in}t\ln \left( {1 + \frac{3}{t}} \right) \to \left( \infty \right)\left( 0 \right)$$

which is an indeterminate form, but not one that allows us to use L'Hôpital's Rule. As discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L'Hôpital's Rule to be applied.

The real question is do we move the first term or the second term to the denominator. From the looks of things, it appears that it would be best to move the first term to the denominator.

$$\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \frac{3}{t}} \right)} \right] = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}}$$

Notice as well that,

$${\mbox{as}}\,\,\,t \to \infty \hspace{0.5in}\frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}} \to \frac{0}{0}$$

and we can use L'Hôpital's Rule on this. Applying L'Hôpital's Rule gives,

$$\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \frac{3}{t}} \right)} \right] = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}} = \mathop {\lim }\limits_{t \to \infty } \frac{{\frac{{ - {}^{3}/{}_{{{t^2}}}}}{{1 + {}^{3}/{}_{t}}}}}{{ - {}^{1}/{}_{{{t^2}}}}}.$$

Can you see why we chose to move the $t$ to the denominator? Moving the logarithm would have left us with a very messy derivative to take! It might have ended up working okay for us, but the work would be greatly increased.

We can't forget to simplify the expression now, as it will greatly help us determine the limit.

$$\frac{{\frac{{ - {}^{3}/{}_{{{t^2}}}}}{{1 + {}^{3}/{}_{t}}}}}{{ - {}^{1}/{}_{{{t^2}}}}}=\frac{3}{1+{}^3/{}_t}$$

Trying the limit now, we have

$$\mathop {\lim }\limits_{t \to \infty } \left[ {t\ln \left( {1 + \frac{3}{t}} \right)} \right] = \mathop {\lim }\limits_{t \to \infty } \frac{{\ln \left( {1 + \frac{3}{t}} \right)}}{{{}^{1}/{}_{t}}} = \mathop {\lim }\limits_{t \to \infty } \frac{3}{{1 + {}^{3}/{}_{t}}} = 3$$

and so we are done!



8. $\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right]$
Click for solution


Let's begin by direct substitution. A quick check shows us that,

$${\mbox{as}}\,\,\,w \to {0^ + }\hspace{0.5in}{w^2}\ln \left( {4{w^2}} \right) \to \left( 0 \right)\left( { - \infty } \right)$$

which is an indeterminate form, but not one that allows us to use L'Hôpital's Rule. As discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L'Hôpital's Rule to be applied.

The real question is do we move the first term or the second term to the denominator. From the looks of things, it appears that it would be best to move the first term to the denominator.

$$\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right] = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}}$$

Notice as well that,

$${\mbox{as}}\,\,\,w \to {0^ + }\hspace{0.5in}\frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}} \to \frac{{ - \infty }}{\infty }$$

and we can use L'Hôpital's Rule on this. Applying L'Hôpital's Rule gives,

$$\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right] = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}} = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{{}^{2}/{}_{w}}}{{ - {}^{2}/{}_{{{w^3}}}}}.$$

Can you see why we chose to move the $t$ to the denominator? Moving the logarithm would have left us with a very messy derivative to take! It might have ended up working okay for us, but the work would be greatly increased.

We can't forget to simplify the expression now, as it will greatly help us determine the limit.

$$\mathop {\lim }\limits_{w \to {0^ + }} \left[ {{w^2}\ln \left( {4{w^2}} \right)} \right] = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{\ln \left( {4{w^2}} \right)}}{{{}^{1}/{}_{{{w^2}}}}} = \mathop {\lim }\limits_{w \to {0^ + }} \frac{{{}^{2}/{}_{w}}}{{ - {}^{2}/{}_{{{w^3}}}}} = \mathop {\lim }\limits_{w \to {0^ + }} \left( { - {w^2}} \right) = 0$$


9. $\mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {x - 1} \right)\tan \left( {\displaystyle \frac{\pi }{2}x} \right)} \right]$
Click for solution


Let's begin by direct substitution. A quick check shows us that,

$${\mbox{as}}\,\,\,x \to {1^ + }\hspace{0.5in}\left( {x - 1} \right)\tan \left( {\frac{\pi }{2}x} \right) \to \left( 0 \right)\left( { - \infty } \right)$$

which is an indeterminate form, but not one that allows us to use L'Hôpital's Rule. As discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L'Hôpital's Rule to be applied.

The real question is do we move the first term or the second term to the denominator. At first glance it might appear that neither term will be particularly useful in the denominator. In particular, if we move the tangent to the denominator we would end up needing to differentiate a term in the form $1/\tan(\theta)$. That doesn’t look to be all that fun to differentiate and we’re liable to end up with a mess when we are done.

However, that is exactly the term we are going to move to the denominator for reasons that will quickly become apparent.

$$\mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {x - 1} \right)\tan \left( {\frac{\pi }{2}x} \right)} \right] = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1}}{{{}^{1}/{}_{{\tan \left( {\frac{\pi }{2}x} \right)}}}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1}}{{\cot \left( {\frac{\pi }{2}x} \right)}}$$

With a little simplification after moving the tangent to the denominator we ended up with something that doesn’t look all that bad. We’ll also see that the remainder of this problem is going to be quite simple.

Before we proceed however we should notice as well that,

$${\mbox{as}}\,\,\,x \to {1^ + }\hspace{0.5in}\frac{{x - 1}}{{\cot \left( {\frac{\pi }{2}x} \right)}} \to \frac{0}{0}$$

and we can use L'Hôpital's Rule on this. Applying L'Hôpital's Rule gives,

$$\mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {x - 1} \right)\tan \left( {\frac{\pi }{2}x} \right)} \right] = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1}}{{\cot \left( {\frac{\pi }{2}x} \right)}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{ - \frac{\pi }{2}{{\csc }^2}\left( {\frac{\pi }{2}x} \right)}} = - \frac{2}{\pi }$$

and we are done!



10. $\mathop {\lim }\limits_{y \to {0^ + }} {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}}$
Click for solution


Let's begin by direct substitution. A quick check shows us that,

$${\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}{\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}} \to {1^\infty }$$

which is an indeterminate form, but not one that allows us to use L'Hôpital's Rule. As discussed in the notes for this section we can do some manipulation on this to turn it into a problem that can be done with L'Hôpital's Rule.

First, let’s define,

$$z = {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}}$$

and take the natural logarithm of both sides. We’ll also do a little simplification.

$$\ln z = \ln \left( {{{\left[ {\cos \left( {2y} \right)} \right]}^{{}^{1}/{}_{{{y^{\,2}}}}}}} \right) = \frac{1}{{{y^{\,2}}}}\ln \left[ {\cos \left( {2y} \right)} \right] = \frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}$$

We can now take the limit as $y\to0^+$ of this.

$$\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}} \right]$$

Before we proceed let’s notice that we have the following,

$${\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}} \to \frac{{\ln \left( 1 \right)}}{0} = \frac{0}{0}$$

and we can use L'Hôpital's Rule on this. Applying L'Hôpital's Rule gives,

$$\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \left[ {\frac{{\ln \left[ {\cos \left( {2y} \right)} \right]}}{{{y^2}}}} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{{}^{{ - 2\sin \left( {2y} \right)}}/{}_{{\cos \left( {2y} \right)}}}}{{2y}} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - \tan \left( {2y} \right)}}{y}.$$

We now have a limit that behaves like,

$${\mbox{as}}\,\,\,y \to {0^ + }\hspace{0.5in}\frac{{ - \tan \left( {2y} \right)}}{y} \to \frac{0}{0}$$

and so we can use L'Hôpital's Rule on this as well. Doing this gives,

$$\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right] = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - \tan \left( {2y} \right)}}{y} = \mathop {\lim }\limits_{y \to {0^ + }} \frac{{ - 2{{\sec }^2}\left( {2y} \right)}}{1} = - 2.$$

Now all we need to do is recall that,

$$z=e^{\ln(z)}.$$

This in turn means that we can do the original limit as follows,

$$\mathop {\lim }\limits_{y \to {0^ + }} {\left[ {\cos \left( {2y} \right)} \right]^{{}^{1}/{}_{{{y^{\,2}}}}}} = \mathop {\lim }\limits_{y \to {0^ + }} z = \mathop {\lim }\limits_{y \to {0^ + }} {{\bf{e}}^{\ln z}} = {{\bf{e}}^{\mathop {\lim }\limits_{y \to {0^ + }} \left[ {\ln z} \right]}} = {{\bf{e}}^{ - 2}}$$

and we are done!



11. $\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}$
Click for solution


Let's begin by seeing what happens as $x\to\infty$. A quick check shows us that,

$${\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}{\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} \to {\infty ^0}$$

which is an indeterminate form, but not one that allows us to use L'Hôpital's Rule. As discussed in the notes for this section we can do some manipulation on this to turn it into a problem that can be done with L'Hôpital's Rule.

First, let’s define,

$$z = {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}}$$

and take the natural logarithm of both sides. We’ll also do a little simplification.

$$\ln z = \ln \left( {{{\left[ {{{\bf{e}}^x} + x} \right]}^{{}^{1}/{}_{x}}}} \right) = \frac{1}{x}\ln \left[ {{{\bf{e}}^x} + x} \right] = \frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}$$

We can now take the limit as $x\to\infty$ of this.

$$\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right]$$

Before we proceed let’s notice that we have the following,

$${\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x} \to = \frac{\infty }{\infty }$$

and we can use L'Hôpital's Rule on this. Applying L'Hôpital's Rule gives,

$$\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\ln \left[ {{{\bf{e}}^x} + x} \right]}}{x}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{}^{{{{\bf{e}}^x} + 1}}/{}_{{{{\bf{e}}^x} + x}}}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}}.$$

We now have a limit that behaves like,

$${\mbox{as}}\,\,\,x \to \infty \hspace{0.5in}\frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} \to \frac{\infty }{\infty }$$

and so we can use L'Hôpital's Rule on this as well. Doing this gives,

$$\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x} + 1}}{{{{\bf{e}}^x} + x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x} + 1}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\bf{e}}^x}}}{{{{\bf{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \left( 1 \right) = 1$$

Notice that we did have to use L'Hôpital's Rule twice here and we also made sure to do some simplification so we could actually take the limit.

Now all we need to do is recall that,

$$z = {{\bf{e}}^{\ln z}}$$

This in turn means that we can do the original limit as follows,

$$\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^x} + x} \right]^{{}^{1}/{}_{x}}} = \mathop {\lim }\limits_{x \to \infty } z = \mathop {\lim }\limits_{x \to \infty } {{\bf{e}}^{\ln z}} = {{\bf{e}}^{\mathop {\lim }\limits_{x \to \infty } \left[ {\ln z} \right]}} = \bf{e}$$

and we are done!



12. $\displaystyle\lim_{x\to\infty}\dfrac{2x+\sqrt{x^2-7}}{4-5x}$
Click for solution


Let's begin by seeing what happens as $x\to\infty$. A quick check shows us that,

$$\displaystyle\lim_{x\to\infty}\dfrac{2x+\sqrt{x^2-7}}{4-5x}\to\dfrac{\infty}{-\infty}$$

which is an indeterminate form that allows us to use L'Hôpital's Rule. So let's do that. Keep in mind we'll have to use the chain rule for the square root in the numerator. We'll also be sure to simplify our expression after applying L'Hôpital's Rule.

$$\begin{align*} \lim_{x\to\infty}\dfrac{2x+\sqrt{x^2-7}}{4-5x} &= \lim_{x\to\infty}\dfrac{2+\frac{1}{2}(x^2-7)^{-1/2}(2x)}{-5} && \text{apply L'Hôpital's Rule} \\ &= \lim_{x\to\infty}\dfrac{2\sqrt{x^2-7}+x}{-5\sqrt{x^2-7}} && \text{simplify} \end{align*}$$

A quick check shows that this new limit also goes to $\frac{\infty}{-\infty}$ as $x\to\infty$. Ok, no problem. We've seen before that we may need to use L'Hôpital's Rule more than once (as long as we still meet the conditions to do so). So let's apply L'Hôpital's Rule again, as well as simplifying the expression.

$$\begin{align*} \lim_{x\to\infty}\dfrac{2x+\sqrt{x^2-7}}{4-5x} &= \lim_{x\to\infty}\dfrac{2\sqrt{x^2-7}+x}{-5\sqrt{x^2-7}} && \text{from before} \\ &= \lim_{x\to\infty}\dfrac{2\cdot\frac{1}{2}(x^2-7)^{-1/2}(2x)+1}{-5\cdot\frac{1}{2}(x^2-7)^{-1/2}(2x)} && \text{apply L'Hôpital's Rule} \\ &= \lim_{x\to\infty}\dfrac{2x+\sqrt{x^2-7}}{-5x} && \text{simplify} \end{align*}$$

While this new limit would also yeild $\frac{\infty}{-\infty}$, the bigger concern is that this is practically the same limit we started out with. What this tells us is that applying L'Hôpital's Rule (even though we're "allowed" to), is not going to be helpful. We'll just be stuck in a cycle if we tried to repeatedly apply L'Hôpital's Rule.

In cases like these, we need to go back to our algebraic methods from before. Here, we can factor out the largest power of $x$ and see if the resulting limit is easier to compute.

We'll want to start with factoring on the inside of the root, and then continue with factoring from the numerator and denominator.

$$\begin{align*} \lim_{x\to\infty}\dfrac{2x+\sqrt{x^2-7}}{4-5x} &= \lim_{x\to\infty}\dfrac{2x+\sqrt{x^2\left(1-\frac{7}{x^2}\right)}}{4-5x} && \text{factor $x^2$ on inside of root} \\ &= \lim_{x\to\infty}\dfrac{2x+x\sqrt{1-\frac{7}{x^2}}}{4-5x} && \text{$\sqrt{x^2}=x$ since $x\to\infty$}\\ &= \lim_{x\to\infty}\dfrac{x\left(2+\sqrt{1-\frac{7}{x^2}}\right)}{x\left(\frac{4}{x}-5\right)} && \text{factor $x$ from numerator and denominator} \\ &= \lim_{x\to\infty}\dfrac{2+\sqrt{1-\frac{7}{x^2}}}{\frac{4}{x}-5} && \text{cancel common factor} \\ &= \dfrac{2+\sqrt{1-0}}{0-5} && \text{apply limit: $\frac{7}{x^2}\to0$ and $\frac{4}{x}\to0$ as $x\to\infty$} \\ &= -\dfrac{2}{5} && \text{simplify} \end{align*}$$

And so the value of our limit is $-\frac{2}{5}$. This problem is a reminder that we do not have to use L'Hôpital's Rule when our algebraic methods will work. And in this situation, L'Hôpital's Rule didn't help anyways.



13. $\displaystyle\lim_{x\to\infty}\left(\ln(x)-x^2\right)$
Click for solution


Let's begin by seeing what happens as $x\to\infty$. A quick check shows us that,

$$\displaystyle\lim_{x\to\infty}\left(\ln(x)-x^2\right)\to\infty-\infty$$

which is an indeterminate form, but not one that allows us to use L'Hôpital's Rule. So we'll have to so some algebraic manipulation to rewrite it. As we had done with polynomials, let's factor out $x^2$ from the entire expression.

$$\displaystyle\lim_{x\to\infty}\left(\ln(x)-x^2\right)=\displaystyle\lim_{x\to\infty}x^2\left(\dfrac{\ln(x)}{x^2}-1\right)$$

Let's just look at the expression $\dfrac{\ln(x)}{x^2}$ for a moment. As $x\to\infty$, $\dfrac{\ln(x)}{x^2}\to\dfrac{\infty}{\infty}$. So we can apply L'Hôpital's Rule to this particular limit. Let's do this.

$$\begin{align*} \lim_{x\to\infty}\dfrac{\ln(x)}{x^2} &= \lim_{x\to\infty}\dfrac{1/x}{2x} && \text{apply L'Hôpital's Rule} \\ &= \lim_{x\to\infty}\dfrac{1}{2x^2} && \text{simplify} \\ &= 0 && \text{apply limit} \end{align*}$$

Keep in mind that this was not the limit we were asked about, but only part of an overall limit. But we can use what we found here to help determine the overall limit.

$$\begin{align*} \lim_{x\to\infty}x^2\left(\dfrac{\ln(x)}{x^2}-1\right) &= \left(\lim_{x\to\infty}x^2\right)\left(\lim_{x\to\infty}\dfrac{\ln(x)}{x^2}-\lim_{x\to\infty}1\right) && \text{properties of limits} \\ &= \infty(0-1) && \text{apply limit; use what we found above} \\ &= -\infty && \text{simplify} \end{align*}$$

So we see the overall limit tends to $-\infty$ (which does not exist) as $x\to\infty$.



14. $\displaystyle\lim_{x\to\infty}\left(e^x-x^3\right)$
Click for solution


Let's begin by seeing what happens as $x\to\infty$. A quick check shows us that,

$$\lim_{x\to\infty}\left(e^x-x^3\right)\to\infty-\infty$$

which is an indeterminate form, but not one that allows us to use L'Hôpital's Rule. So we'll have to so some algebraic manipulation to rewrite it. As we had done with polynomials, let's factor out $x^3$ from the entire expression.

$$\lim_{x\to\infty}\left(e^x-x^3\right)=\displaystyle\lim_{x\to\infty}x^3\left(\dfrac{e^x}{x^3}-1\right)$$

Let's just look at the expression $\dfrac{e^x}{x^3}$ for a moment. As $x\to\infty$, $\dfrac{e^x}{x^3}\to\dfrac{\infty}{\infty}$. So we can apply L'Hôpital's Rule to this particular limit. Let's do this.

$$\begin{align*} \lim_{x\to\infty}\dfrac{e^x}{x^3} &= \lim_{x\to\infty}\dfrac{e^x}{3x^2} && \text{apply L'Hôpital's Rule; still goes to $\frac{\infty}{\infty}$} \\ &= \lim_{x\to\infty}\dfrac{e^x}{6x} && \text{apply L'Hôpital's Rule again; still goes to $\frac{\infty}{\infty}$} \\ &= \lim_{x\to\infty}\dfrac{e^x}{6} && \text{apply L'Hôpital's Rule again} \\ &= \frac{\infty}{6} && \text{apply limit} \\ &= \infty && \text{simplify} \end{align*}$$

Keep in mind that this was not the limit we were asked about, but only part of an overall limit. But we can use what we found here to help determine the overall limit.

$$\begin{align*} \lim_{x\to\infty}x^3\left(\dfrac{e^x}{x^3}-1\right) &= \left(\lim_{x\to\infty}x^3\right)\left(\lim_{x\to\infty}\dfrac{e^x}{x^3}-\lim_{x\to\infty}1\right) && \text{properties of limits} \\ &= \infty(\infty-1) && \text{apply limit; use what we found above} \\ &= \infty && \text{simplify} \end{align*}$$

So we see the overall limit tends to $\infty$ (which does not exist) as $x\to\infty$.

Assignment Problems

For problems 1 – 18 determine the limit.

$\displaystyle \mathop {\lim }\limits_{x \to - 4} \frac{{{x^3} + 6{x^2} - 32}}{{{x^3} + 5{x^2} + 4x}}$
$\displaystyle \mathop {\lim }\limits_{w \to \, - \infty } \frac{{{{\bf{e}}^{ - 6w}}}}{{4 + {{\bf{e}}^{ - 3w}}}}$
$\displaystyle \mathop {\lim }\limits_{t \to 0} \frac{{\sin \left( {6t} \right)}}{{\sin \left( {11t} \right)}}$
$\displaystyle \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 8x - 9}}{{{x^3} - 2{x^2} - 5x + 6}}$
$\displaystyle \mathop {\lim }\limits_{t \to 2} \frac{{{t^3} - 7{t^2} + 16t - 12}}{{{t^4} - 4{t^3} + 4{t^2}}}$
$\displaystyle \mathop {\lim }\limits_{w \to - \infty } \frac{{{w^2} - 4w + 1}}{{3{w^2} + 7w - 4}}$
$\displaystyle \mathop {\lim }\limits_{y \to \infty } \frac{{{y^2} - {{\bf{e}}^{6\,y}}}}{{4{y^2} + {{\bf{e}}^{7\,y}}}}$
$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{2\cos \left( {4x} \right) - 4{x^2} - 2}}{{\sin \left( {2x} \right) - {x^2} - 2x}}$
$\displaystyle \mathop {\lim }\limits_{x \to \, - 3} \frac{{3{{\bf{e}}^{2\,x + 6}} + {x^2} - 12}}{{{x^3} + 6{x^2} + 9x}}$
$\displaystyle \mathop {\lim }\limits_{z \to 6} \frac{{\sin \left( {\pi z} \right)}}{{\ln \left( {z - 5} \right)}}$
$\mathop {\lim }\limits_{w \to \infty } \left[ {w\ln \left( {1 - \displaystyle \frac{2}{{3w}}} \right)} \right]$
$\mathop {\lim }\limits_{t \to {0^ + }} \left[ {\ln \left( t \right)\sin \left( t \right)} \right]$
$\mathop {\lim }\limits_{z \to - \infty } {z^2}{{\bf{e}}^z}$
$\displaystyle\lim_{x\to-\infty}\dfrac{3x+5}{\sqrt{9+x^2}}$
$\mathop {\lim }\limits_{x \to \infty } \left[ {x\sin \left( {\displaystyle \frac{7}{x}} \right)} \right]$
$\mathop {\lim }\limits_{z \to {0^ + }} \left[ {{z^2}{{\left( {\ln z} \right)}^2}} \right]$
$\mathop {\lim }\limits_{x \to {0^ + }} {x^{{}^{1}/{}_{x}}}$
$\displaystyle\lim_{x\to\infty}\left(\ln(x)-\sqrt{x}\right)$
$\mathop {\lim }\limits_{t \to {0^ + }} {\left[ {{{\bf{e}}^t} + t} \right]^{{}^{1}/{}_{t}}}$
$\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\bf{e}}^{ - 2x}} + 3x} \right]^{{}^{1}/{}_{x}}}$
1. Suppose that we know that $f'(x)$ is a continuous function. Use L'Hôpital's Rule to show that, $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( {x - h} \right)}}{{2h}} = f'\left( x \right).$$
2. Suppose that we know that $f''(x)$ is a continuous function. Use L'Hôpital's Rule to show that, $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - 2f\left( x \right) + f\left( {x - h} \right)}}{{{h^2}}} = f''\left( x \right).$$


3. Suppose you that $f(10) = -9, f'(10) = -8$, and $f'(x)$ is continuous for all $x$. Find the following.
1. $\lim\limits_{x\to 10}\frac{f(x)+9}{x^2-100}$


2. $\lim\limits_{x\to 10}\frac{\ln(f(x)+10)}{x-10}$


4. Suppose you that $f(-6) = 7, f'(-6) = -8$, and $f'(x)$ is continuous for all $x$. Find the following.
1. $\lim\limits_{x\to -6}\frac{\ln(f(x)-6)}{[f(x)]^2-49}$

Note: The following content is adapted from OpenStax and is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction.

Evalute the following limits.

25. $\lim\limits_{x\to 0}\frac{\sin(x)-\tan(x)}{x^3}$


26. $\lim\limits_{x\to 0}\frac{(1+x)^n - 1}{x}$


27. $\lim\limits_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$


28. $\lim\limits_{x\to0^+}x\ln(x^4)$


29. $\lim\limits_{x\to0}\frac{3^x-2^x}{x}$


30. $\lim\limits_{x\to\infty}xe^{1/x}$


31. $\lim\limits_{x\to0^+}x^{1/\cos(x)}$