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### Course: Differential Calculus > Unit 4

Lesson 8: L’Hôpital’s rule: composite exponential functions# L'Hôpital's rule review

L'Hôpital's rule helps us find many limits where direct substitution ends with the indeterminate forms 0/0 or ∞/∞. Review how (and when) it's applied.

## What is L'Hôpital's rule?

L'Hôpital's rule helps us evaluate indeterminate limits of the form $\frac{0}{0}$ or $\frac{\mathrm{\infty}}{\mathrm{\infty}}$ .

In other words, it helps us find $\underset{x\to c}{lim}{\displaystyle \frac{u(x)}{v(x)}}$ , where $\underset{x\to c}{lim}u(x)=\underset{x\to c}{lim}v(x)=0$ (or, alternatively, where both limits are $\pm \mathrm{\infty}$ ).

The rule essentially says that $\underset{x\to c}{lim}{\displaystyle \frac{{u}^{\prime}(x)}{{v}^{\prime}(x)}}$

*if*the limit*exists*, then the two limits are equal:*Want to learn more about L'Hôpital's rule? Check out this video.*

## Using L'Hôpital's rule to find limits of quotients

Let's find, for example, $\underset{x\to 0}{lim}{\displaystyle \frac{7x-\mathrm{sin}(x)}{{x}^{2}+\mathrm{sin}(3x)}}$ .

Substituting $x=0$ into $\frac{7x-\mathrm{sin}(x)}{{x}^{2}+\mathrm{sin}(3x)}$ results in the indeterminate form $\frac{0}{0}$ . So let's use L’Hôpital’s rule.

Note that we were only able to use L’Hôpital’s rule because the limit $\underset{x\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{d}{dx}}[7x-\mathrm{sin}(x)]}{{\displaystyle \frac{d}{dx}}[{x}^{2}+\mathrm{sin}(3x)]}}$ actually exists.

*Want to try more problems like this? Check out this exercise.*

## Using L'Hôpital's rule to find limits of exponents

Let's find, for example, $\underset{x\to 0}{lim}(1+2x{)}^{{}^{\frac{1}{\mathrm{sin}(x)}}}$ . Substituting $x=0$ into the expression results in the indeterminate form ${1}^{{}^{\mathrm{\infty}}}$ .

To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(1+2x{)}^{{}^{\frac{1}{\mathrm{sin}(x)}}}$ , we will find $\underset{x\to 0}{lim}\mathrm{ln}(y)$ . Once we find it, we will be able to find $\underset{x\to 0}{lim}y$ .

Substituting $x=0$ into $\frac{\mathrm{ln}(1+2x)}{\mathrm{sin}(x)}$ results in the indeterminate form $\frac{0}{0}$ , so now it's L’Hôpital’s rule's turn to help us with our quest!

We found that $\underset{x\to 0}{lim}\mathrm{ln}(y)=2$ , which means $\underset{x\to 0}{lim}y={e}^{2}$ .

*Want to try more problems like this? Check out this exercise.*

## Want to join the conversation?

- Why is 1^infinity an indeterminate form?(4 votes)
- This stems from the fact that all of the limits in calculus of this type have something to do with the number
**e**(2.71828...)

e is actually defined as limit(n->infinity, (1+1/n)^n).

At first glance, (1+1/n) seems to be 1 and hence, this is called 1^infinity form. However, the limit of this quantity is 2.718...(5 votes)

- In the article's example for using L'Hopital's rule for finding limits of exponents, they get (1+2(0))^1/sin(0) = 1^infinity (direct substitution). But won't 1/sin(0) be undefined, thus resulting in 1^undefined = undefined?(4 votes)
- In the explanation to problem 1.2, the derivative of the top equation is shown as

.`d/dx[xcos(πx)] = cos(xπ) - πxsin(xπ)`

Where does the`π`

in`- πxsin`

... come from?(1 vote)- We multiply by π because (by the chain rule) we're multiplying by the derivative of xπ, the function inside of the sine function.(4 votes)

- what is the derivative of e^1/2 or e^0.5(1 vote)
- e^0.5 is a constant, around 1.65. So the derivative is just 0.(2 votes)

- i watched the video.(1 vote)
- in the exercises, I was confused

how: ln(y) =(x-1)ln(x-1)

became: ln(y)=( ln(x-1) ) / (x-1)^-1

could someone help?

the question was to find the limit as x approaches 1 from the positive side for (x-1)^(x-1)(1 vote)- (x-1) is the same as 1/(x-1)^-1. Putting it in that form makes it useful for checking L'Hopital's Rule because we don't care about the overall function/quotient to start with, just the individual functions themselves.

The limit of (x-1)^-1 = 1/(x-1) as x approaches 1 from the positive direction is infinity, which is readily apparent if you graph it on Desmos or some such. Going through L'Hopital's rule you'll eventually get the limit of Ln(y) = 0, so for that to be true the limit of y, and thus the limit we want, must be 1. Hope that helps.(1 vote)

- Is there a printable version of this page?(1 vote)
- Unfortunately there is nothing like that. So far best solution might be using Snipping tool on windows which can easily cut "pictures" from browser and then you can arrange them together in some software( Microsoft word would suffice).(1 vote)

- I had a problem (1-4/x)^x . My question was when they took (4/x^2)/((1-4/x)(-1/x^2)) and got (4x^2)/(1-4/x)(-x^-2).(1 vote)
- When using L'Hôpital's rule to find limits of exponents, there's a step that sets, for example, lim x->∞ ln(y) equal to ln (lim x->∞ y). Which logarithm or limit property allows this?(1 vote)
- Here we can use this property because here we are not applying the limit to whole ln(y(x)) operator we have our variable x in the y(x) , So here we just wanna find the limiting value of y(x)

It doesn't violate our previous method that we use we just plug the value and try to come up w/ a more subtle and concrete way of understanding this.(1 vote)

- What about lim x→0 cot(x)/In(x)?If you apply L'Hôpital's rule,try to differentiate this,you will get into great trouble!

(0 votes)- Direct substitution gives ∞/∞, so taking the derivatives of according to l'Hôpital yields

-csc²(x)/(1/x). This rearranges into -x/sin²(x).

Direct substitution now yields 0/0, so we can apply l'Hôpital's Rule again. Differentiate to get -1/(2sin(x)cos(x))

Now, finally, direct substitution yields -1/0, which indicates that the limit does not exist.(2 votes)