## L'Hôpital's Rule

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### Video List

l'hôpital's rule

A13  A14  A15  A16  indeterminate quotients

B07  B08  indeterminate products

A17  B09  B10  indeterminate differences

B11  B12  indeterminate powers

intro to l'hôpital's rule

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Alternate Names For L'Hôpital's Rule

L'Hospital's Rule

Bernoulli's Rule

#### L'Hôpital's Rule

If the limit $$\displaystyle{ \lim_{x \to c}{\frac{n(x)}{d(x)}} }$$ is indeterminate of the type $$0/0$$ then

$$\displaystyle{ \lim_{x \to c}{\frac{n(x)}{d(x)}} = \lim_{x \to c}{\frac{n'(x)}{d'(x)}} }$$.

### Quick Notes

 This rule can be repeated until a determinate form is found. $$x$$ can approach a finite value $$c$$, $$\infty$$ or $$-\infty$$ For details about determinate and indeterminate forms, click here. This is not the quotient rule. For information about the difference, click here. Do not use this on determinate forms. It will often give you the wrong answer.

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L'Hôpital's Rule is used to evaluate a limit when other techniques will not work, like factoring and rationalizing. It especially convenient to use with exponentials and, sometimes, limits involving trig functions. In fact, L'Hôpital's Rule can usually be used on limits where those other techniques work and it is often easier than factoring. So it is a nice tool to have when evaluating limits.

Some Things To Notice

1. Be careful to use L'Hôpital's Rule only on limits of indeterminate form. If you use it on a determinate form, you may ( and probably will ) get an incorrect answer.
2. L'Hôpital's Rule can be used multiple times. So, if it doesn't work the first time, check that you still have an indeterminate form, and then use it again.
3. One of the big mistakes that students make when they are first learning L'Hôpital's Rule is to confuse it with the derivative of the function itself. Read the next panel to clarify it in your mind.

### Difference Between L'Hôpital's Rule and the Quotient Rule

It is very easy to confuse L'Hôpital's Rule with the Quotient Rule since they are so similar. Here is a rundown of the two for comparison.

1. Quotient Rule - The quotient rule is used on a quotient or ratio of terms to calculate the derivative of a function. The result of the quotient rule is the slope of the original function at all points along the curve. So you can use the result to determine the slope, calculate the equation of a tangent line, find extrema and continue on to determine the second derivative, to name a few uses.
To calculate the quotient rule, we use the following equations. Given that we have a function in the form $$\displaystyle{ f(x) = \frac{n(x)}{d(x)} }$$, the derivative $$f(x)$$ using the quotient rule is $$\displaystyle{ f'(x) = \frac{d(x)n'(x) - n(x)d'(x)}{[d(x)]^2} }$$

2. L'Hôpital's Rule - L'Hôpital's Rule is used only when we want to calculate a limit and can be used only under very specific circumstances (see the discussion below for a complete explanation). L'Hôpital's Rule works like this.
If we have a limit that goes to an indeterminate form, for example $$\displaystyle{ \lim_{x \to c}{\left[ \frac{n(x)}{d(x)} \right]} }$$ where $$\displaystyle{ \lim_{x \to c}{[n(x)]} = 0 }$$ and $$\displaystyle{ \lim_{x \to c}{[d(x)]} = 0 }$$ giving us $$0/0$$ in the original limit, then L'Hôpital's Rule tells us that $$\displaystyle{ \lim_{x \to c}{\frac{n(x)}{d(x)}} = \lim_{x \to c}{\frac{n'(x)}{d'(x)}} }$$.

Okay, now let's compare the two using $$\displaystyle{ f(x) = \frac{n(x)}{d(x)} }$$

Quotient Rule L'Hôpital's Rule Context: Derivatives Context: Limits One of several ways to calculate the derivative of a function. Can be used only on indeterminate limits. The resulting equation has many uses including calculating the slope of a curve. The resulting fraction $$[n'(x)/d'(x)]$$ has no meaning other than its limit is the same as the limit of the original function.

Study Note - When you are learning two similar concepts or you are learning a new concept that is similar to one you already know and you are getting confused, try the technique we used above. This will help you separate them in your mind and know when to use which technique. This also works for learning similar words in foreign languages. Click here for more study techniques.

Indeterminate Forms

Note - - You do not need to understand L'Hôpital's Rule to learn about indeterminate forms here.
The word 'indeterminate' means unable to determine. When you have an indeterminate form the value cannot be determinated. For example, $$0/0$$ can mean $$1$$ or $$-1227$$ or $$-\infty$$ or almost any number. There is no way to tell without changing the form (fascinating, isn't it?). That's where L'Hôpital's Rule comes in. L'Hôpital's Rule allows us to change the form into something else so that we CAN determine the value we are looking for when it comes to limits.

The indeterminate forms table below shows most of the indeterminate forms that you will run across as well as some determinate forms to watch for. However, you do not need to memorize these forms to determine if you need to use L'Hôpital's Rule. There is an easier way. All you need to remember is that indeterminate form is $$0/0$$ and determinate form is $$c/0 = \pm \infty$$ where $$c \neq 0$$. Then use a little bit of algebra to get one of these forms. Let's do some examples.

Determine if $$\displaystyle{ \infty / \infty }$$ is determinate or indeterminate.

Determine if $$1^{\infty}$$ is determinate or indeterminate.

Now that you have some guidelines, try showing these on the other items in the indeterminate forms table. This will not only give you practice, but it will help you remember the techniques quickly at exam time.

Indeterminate Forms

Determinate Forms

$$0/0$$ $$\infty + \infty = \infty$$
$$\pm \infty / \pm \infty$$ $$- \infty - \infty = - \infty$$
$$\infty - \infty$$ $$0^{\infty} = 0$$
$$0 (\infty)$$ $$0^{-\infty} = \infty$$
$$0^0$$ $$(\infty) \cdot (\infty) = \infty$$
$$1^{\infty}$$
$$\infty ^ 0$$

Use L'Hôpital's Rule

Do Not Use L'Hôpital's Rule

Types of Indeterminate Forms

From the discussion so far, it looks like we have a lot going on and there are a lot of cases in the indeterminate forms table. However, there are really only four cases and only two techniques you need to cover all the cases. Let's look at the cases.

 Case 1 - - Indeterminate Quotients In a perfect world, all your problems would be in this form because you can just apply L'Hôpital's Rule directly. These are problems where you already have a limit in the form $$\displaystyle{ \lim_{x \to a}{\frac{n(x)}{d(x)}} }$$ And, when you plug in $$x=a$$, you get either $$0/0$$ or $$\pm \infty / \pm \infty$$. In the last indeterminate form with $$\infty$$, the signs can be anything, so you have 4 possible cases. In any case, your problem is all set up to use L'Hôpital's Rule and you just directly apply the rule. Case 2 - - Indeterminate Products In this case, you have an determinate form that looks like $$0(\infty)$$ or $$0(-\infty)$$, in which case you just need to do a little bit of algebra to convert one of the pieces of the product into a denominator and you end up with case 1. This sometimes takes some thinking. For example, if you have $$\displaystyle{ \lim_{x \to 0^+}{x\ln(x)} }$$, you can rewrite $$x$$ as $$1/x^{-1}$$ and the limit now is $$\displaystyle{ \lim_{x \to 0^+}{\ln(x)/x^{-1}} }$$ where $$n(x) = \ln(x)$$ and $$d(x) = x^{-1}$$ and you have the indeterminate form $$\infty / \infty$$ and you can use L'Hôpital's Rule. Case 3 - - Indeterminate Differences In this case, you have the indeterminate form that looks like $$\infty - \infty$$ and we need a fraction to be able to apply L'Hôpital's Rule. To get the right form, just find a common a denominator and combine the terms to get an indeterminate form that looks like case 1. Case 4 - - Indeterminate Powers In the previous three cases, you either started with a fraction in the right form (case 1) or did some algebra to get your function into the right form (cases 2 and 3). In this case, you need another technique. The idea is that you have an indeterminate form that looks like $$0^0$$, $$1^{\infty}$$ or $$\infty ^ 0$$ and you need to get some kind of fraction in order to apply L'Hôpital's Rule. To do this we use the natural logarithm. Let's go through a general example. We have the limit $$\displaystyle{ \lim_{x \to a}{[f(x)]^{g(x)}} }$$ and $$[f(a)]^{g(a)}$$ is an indeterminate power. Set $$y = [f(x)]^{g(x)}$$ and take the natural log of both sides. So that $$\ln(y) = \ln( [f(x)]^{g(x)} )$$ and using the power rule for logarithms, we have $$\ln(y) = g(x) \ln[ f(x) ]$$. We take the limit of $$\ln(y)$$ using one cases 1-3 above to get a value $$L$$, which may be finite or infinite. To get the final answer, we 'undo' the natural log by putting $$L$$ in the exponent of $$e$$ giving $$e^L$$. This discussion may be a bit confusing but, if you look at example 2 above, we used this idea to show that $$1^{\infty}$$ was indeterminate (although we didn't undo the natural log at the end since the discussion didn't require it). There are also several practice problems below that will help you.

Okay, now that you know something about L'Hôpital's Rule, under what conditions you can use it and various situations you will encounter, let's watch some videos to help explain how it works and see some examples.

This first video is in-depth and a bit long but is well worth watching. The instructor is easy to listen to and he does a good job of explaining the details of L'Hôpital's Rule as well as giving examples and supporting arguments.

Here is a quick introductory explanation of L'Hôpital's Rule. This is good to watch if you are still a little fuzzy on how to use this rule. There is nothing new here but it gives you another perspective.

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### L'Hôpital's Rule Resources

Practice Problems

Instructions - - Evaluate the following limits using L'Hôpital's Rule ( if valid ), giving your answers in exact form.

 Level A - Basic

Practice A01

$$\displaystyle{ \lim_{x \to \pi/2}{\left[ \frac{\tan(2x)}{x-\pi/2} \right]} }$$

Practice A02

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

Practice A03

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

Practice A04

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

##### Practice Problem [ A01 ] Solution

If we plug in $$\pi/2$$ for x, we get $$0/0$$ which is indeterminate. This means we can use L'Hôpital's Rule.

$$\displaystyle{ \lim_{x \to \pi/2}{\left[ \frac{\tan(2x)}{x-\pi/2} \right]} = \lim_{x \to \pi/2}{\left[ \frac{2 \sec^2(2x)}{1} \right]} = 2 \sec^2(\pi) = 2 }$$

Note: You could also solve this using trig identities (but that's a lot more work).

$$\displaystyle{ \lim_{x \to \pi/2}{\left[ \frac{\tan(2x)}{x-\pi/2} \right]} = 2 }$$

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##### Practice Problem [ A02 ] Solution

First of all, you always want to try direct substitution. When you do this, you get $$0/0$$ which is indeterminate. (If you don't know why this is, reread the indeterminate forms section.)
There are at least two ways to solve this.
We will solve it using L'Hôpital's Rule here. See the Finite Limits Page for an alternative solution.
$$\displaystyle{ \lim_{x \rightarrow 2} \left[ \frac{3x^2-x-10}{x^2-4} \right] = \lim_{x \rightarrow 2} \left[ \frac{6x-1}{2x} \right] = \frac{6(2)-1}{2(2)} = 11/4 }$$
Remember: You can use L'Hôpital's Rule only if you have an indeterminate form.

$$\displaystyle{\lim_{x \rightarrow 2}{ \left[ \frac{3x^2-x-10}{x^2-4} \right] }= 11/4 }$$

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##### Practice Problem [ A03 ] Solution

First of all, you always want to try direct substitution. When you do this, you get $$0/0$$ which is indeterminate. (If you don't know why this is, reread the indeterminate forms section.)
There are at least two ways to solve this.
We will solve it using L'Hôpital's Rule here. See the Finite Limits Page for an alternative solution.

$$\displaystyle{ \lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] } = \lim_{x \rightarrow 3}{ \left[\frac{4x^3}{4x-5}\right] } = \frac{4(3)^3}{4(3)-5} = 108/7 }$$

Remember: You can use L'Hôpital's Rule only if you have an indeterminate form.

$$\displaystyle{ \lim_{x \rightarrow 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] }= 108/7 }$$

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##### Practice Problem [ A04 ] Solution

Of course, the first thing we want to try is to plug in $$27$$ directly for $$x$$. This gives us $$0/0$$ which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \begin{array}{rcl} \lim_{x \rightarrow 27}{\left[ \frac{x - 27}{x^{1/3} - 3}\right]} & = & \lim_{x \rightarrow 27}{ \frac{1}{(1/3)x^{-2/3}} } \\ & = & \lim_{x \rightarrow 27}{ \frac{x^{2/3}}{1/3} } \\ & = & \lim_{x \rightarrow 27}{ 3x^{2/3} } \\ & = & 3(27)^{2/3} = 3(9) = 27 \end{array} }$$

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

Note - - We could have used algebra and factoring to solve this problem. See that solution on the Finite Limits Page.

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Practice A05

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

Practice A06

$$\displaystyle{ \lim_{x \to 0}{\left[\frac{x+tan(x)}{sin(x)}\right]} }$$

Practice A07

$$\displaystyle{ \lim_{z \to \pi}{\left[ \frac{z - \pi}{sin(z)} \right]} }$$

Practice A08

$$\displaystyle{ \lim_{y \to 0}{\left[ \frac{sin(3y)}{sin(4y)} \right]} }$$

##### Practice Problem [ A05 ] Solution

Direct substitution yields $$0/0$$, which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \begin{array}{rcl} \lim_{x \rightarrow 1}{\left[ \frac{x^{1/3}-1}{x^{1/4}-1}\right] } & = & \lim_{x \rightarrow 1}{ \frac{(1/3)x^{-2/3}}{(1/4)x^{-3/4}} } \\ & = & \lim_{x \rightarrow 1}{ \frac{4}{3} \frac{x^{3/4}}{x^{2/3}} } \\ & = & \lim_{x \rightarrow 1}{ \frac{4}{3} x^{3/4 - 2/3} } \\ & = & \lim_{x \rightarrow 1}{ \frac{4}{3} x^{9/12 - 8/12} } \\ & = & \lim_{x \rightarrow 1}{ \frac{4}{3} x^{1/12} } = \frac{4}{3} 1^{1/12} = \frac{4}{3} \end{array} }$$

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

Note - - We could have used algebra and factoring to solve this problem. See that solution on the Finite Limits Page.

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##### Practice Problem [ A06 ] Solution

Direct substitution yields $$0/0$$, which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \begin{array}{rcl} \lim_{x \to 0}{\left[\frac{x+tan(x)}{sin(x)}\right]} & = & \lim_{x \to 0}{\left[\frac{1+sec^2(x)}{cos(x)}\right]} \\ & = & \lim_{x \to 0}{\left[\frac{1+[1/cos^2(x)]}{cos(x)}\right]} \\ & = & \lim_{x \to 0}{\left[\frac{cos^2(x)+1}{cos^3(x)}\right]} \\ & = & \frac{1+1}{1} = 2 \end{array} }$$

After applying L'Hôpital's Rule once, the rest is just trig and algebra.

$$\displaystyle{ \lim_{x \to 0}{\left[\frac{x+tan(x)}{sin(x)}\right]} = 2 }$$

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##### Practice Problem [ A07 ] Solution

Direct substitution yields $$0/0$$, which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \lim_{z \to \pi}{\left[ \frac{z - \pi}{sin(z)} \right]} = \lim_{z \to \pi}{\left[ \frac{1}{cos(z)} \right]} = \frac{1}{-1} = -1 }$$

$$\displaystyle{ \lim_{z \to \pi}{\left[ \frac{z - \pi}{sin(z)} \right]} = -1 }$$

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##### Practice Problem [ A08 ] Solution

Direct substitution yields $$0/0$$, which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \lim_{y \to 0}{\left[ \frac{sin(3y)}{sin(4y)} \right]} = \lim_{y \to 0}{\left[ \frac{3cos(3y)}{4cos(4y)} \right]} = \frac{3(1)}{4(1)} = 3/4 }$$

$$\displaystyle{ \lim_{y \to 0}{\left[ \frac{sin(3y)}{sin(4y)} \right]} = 3/4 }$$

Note - - There is another way to work this using the identity $$\displaystyle{ \lim_{x \to 0}{\left[ \frac{sin(x)}{x} \right]} = 1 }$$.

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Practice A09

$$\displaystyle{\lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]} }$$

Practice A10

$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^3 + x + 12}{2x^3 - 5x} \right]} }$$

Practice A11

$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} }$$

Practice A12

$$\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{7x^2 + x + 21}{11-x} \right]} }$$

##### Practice Problem [ A09 ] Solution

Direct substitution yields

$$\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]} = \frac{-\infty}{-\infty}}$$

This is indeterminate. So we can use L'Hôpital's Rule.
$$\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]} = \lim_{x \to -\infty}{ \frac{1}{3} } = 1/3 }$$

$$\displaystyle{\lim_{x \to -\infty}{\left[ \frac{x+5}{3x+7} \right]}= 1/3 }$$

Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the Infinite Limits Page.

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##### Practice Problem [ A10 ] Solution

Direct substitution yields $$\infty - \infty$$ in the denominator, which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^3 + x + 12}{2x^3 - 5x} \right]} = \lim_{x \to \infty}{ \frac{21x^2 + 1}{6x^2} } }$$
Direct substitution still gives us an indeterminate form, $$\infty / \infty$$. So we can use L'Hôpital's Rule again.
$$\displaystyle{ \lim_{x \to \infty}{ \frac{21x^2 + 1}{6x^2} } = \lim_{x \to \infty}{ \frac{42x}{12x} } = \frac{42}{12} = 7/2 }$$

$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^3 + x + 12}{2x^3 - 5x} \right]}= 7/2 }$$

Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the Infinite Limits Page.

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##### Practice Problem [ A11 ] Solution

Direct substitution yields $$\displaystyle{ \frac{\infty - \infty + 12}{\infty + \infty + 127} }$$ which is indeterminate. So we can use L'Hôpital's Rule.
$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} = \lim_{x \to \infty}{\left[ \frac{14x - 3}{3x^2 + 4} \right]} }$$
Again, after direct substitution, we get $$\infty / \infty$$, which is also indeterminate. Use L'Hôpital's Rule again.
$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{14x - 3}{3x^2 + 4} \right]} = \lim_{x \to \infty}{ \frac{14}{6x} } = \frac{14}{\infty} = 0 }$$

$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{7x^2 - 3x + 12}{x^3 + 4x + 127} \right]} = 0 }$$

Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the Infinite Limits Page.

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##### Practice Problem [ A12 ] Solution

Direct substitution yields $$\infty / \infty$$ which is indeterminate. So we can use L'Hôpital's Rule.
$$\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{7x^2 + x + 21}{11-x} \right]} = \lim_{x \to -\infty}{ \frac{14x + 1}{-1} } }$$
Now, let's try direct substitution again.
$$\displaystyle{ \lim_{x \to -\infty}{ \frac{14x + 1}{-1} } = \frac{-\infty + 1}{-1} = \infty }$$

$$\displaystyle{ \lim_{x \to -\infty}{\left[ \frac{7x^2 + x + 21}{11-x} \right]}= +\infty }$$

Note - - This is the easiest way to work this problem. If you don't know L'Hôpital's Rule, you can use algebra. You will find that solution on the Infinite Limits Page.

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Practice A13

$$\displaystyle{ \lim_{x \to 0}{ \frac{sin(x)}{x} } }$$

Practice A14

$$\displaystyle{ \lim_{x \to \infty}{ \frac{ln(x)}{x} } }$$

Practice A15

$$\displaystyle{ \lim_{x \to 0^+}{ \frac{ln(x)}{x} } }$$

Practice A16

$$\displaystyle{ \lim_{x \to \infty}{\frac{x}{ln(1+2e^x)}} }$$

##### Practice Problem [ A13 ] Solution

$$\displaystyle{ \lim_{x \to 0}{\frac{sin(x)}{x}} = 1 }$$

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##### Practice Problem [ A14 ] Solution

$$\displaystyle{ \lim_{x \to \infty}{ \frac{ln(x)}{x} } = 0 }$$

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##### Practice Problem [ A15 ] Solution

This was kind of a trick question to see if you were paying attention.

$$\displaystyle{ \lim_{x \to 0^+}{ \frac{ln(x)}{x} } = -\infty }$$

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##### Practice Problem [ A16 ] Solution

$$\displaystyle{ \lim_{x \to \infty}{\frac{x}{ln(1+2e^x)}} = 1 }$$

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Practice A17

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

##### Practice Problem [ A17 ] Solution

$$\displaystyle{ \lim_{x \to \infty}{(x^2-x)} = \infty }$$

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 Level B - Intermediate

Practice B01

$$\displaystyle{ \lim_{x \to 0}{\left[ \frac{tan(x)-x}{sin(x)-x} \right]} }$$

Practice B02

$$\displaystyle{ \lim_{n \to \infty}{n^{2/n}} }$$

Practice B03

$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^5}{e^{5x}} \right]} }$$

Practice B04

$$\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{k} } }$$ where k is a positive constant

##### Practice Problem [ B01 ] Solution

Direct substitution yields $$0/0$$, which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \begin{array}{rcl} \lim_{x \to 0}{\left[ \frac{tan(x)-x}{sin(x)-x} \right]} & = & \lim_{x \to 0}{\left[ \frac{sec^2(x)-1}{cos(x)-1} \right]} \\ & = & \lim_{x \to 0}{\left[ \frac{2sec(x)[sec(x)tan(x)]}{-sin(x)} \right]} \\ & = & \lim_{x \to 0}{\left[ \frac{2sec^2(x)tan(x)}{-sin(x)} \right]} \\ & = & \lim_{x \to 0}{\left[ \frac{-2sin(x) }{cos^2(x)sin(x)cos(x) } \right]} \\ & = & \lim_{x \to 0}{\left[ \frac{-2}{cos^3(x)} \right]} \\ & = & \frac{-2}{1} = -2 \end{array} }$$

$$\displaystyle{ \lim_{x \to 0}{\left[ \frac{tan(x)-x}{sin(x)-x} \right]} = -2}$$

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##### Practice Problem [ B02 ] Solution

Direct substitution yields $$\infty^0$$ which is indeterminate. We can't directly use L'Hôpital's Rule because L'Hôpital's Rule requires a fraction. So we have to use a trick.

Let's call the limit, $$y$$, i.e. $$\displaystyle{ y = \lim_{n \to \infty}{n^{2/n}} }$$ and take the natural log of both sides. This yields
$$\displaystyle{ \begin{array}{rcl} ln(y) & = & \lim_{n \to \infty}{ln(n^{2/n})} \\ & = & \lim_{n \to \infty}{\frac{2ln(n)}{n}} \\ & = & \lim_{n \to \infty}{\frac{2/n}{1}} = 0 \\ ln(y) & = & 0 \\ e^{ln(y)} & = & e^0 \\ y & = & 1 \end{array} }$$

$$\displaystyle{ \lim_{n \to \infty}{n^{2/n}} = 1 }$$

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##### Practice Problem [ B03 ] Solution

If we try direct substitution first, we get $$\infty/\infty$$ which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \begin{array}{rcl} \lim_{x \to \infty}{\left[ \frac{x^5}{e^{5x}} \right]} & = & \lim_{x \to \infty}{\left[ \frac{5x^4}{5e^{5x}} \right]} \\ & = & \lim_{x \to \infty}{\left[ \frac{x^4}{e^{5x}} \right]} \\ & = & \lim_{x \to \infty}{\left[ \frac{4x^3}{5e^{5x}} \right]} \\ & = & \lim_{x \to \infty}{\left[ \frac{12x^2}{25e^{5x}} \right]} \\ & = & \lim_{x \to \infty}{\left[ \frac{24x}{125e^{5x}} \right]} \\ & = & \lim_{x \to \infty}{\left[ \frac{24}{625e^{5x}} \right]} \\ & = & \frac{24}{\infty} = 0 \end{array} }$$

$$\displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^5}{e^{5x}} \right]} = 0 }$$

Note - - In this solution we used L'Hôpital's Rule five times until we no longer had an indeterminate form. Not shown here is the fact that, before each use of L'Hôpital's Rule, we tested for indeterminate forms by direct substitution.

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##### Practice Problem [ B04 ] Solution

$$\displaystyle{ \begin{array}{rcl} y & = & \lim_{n \to \infty}{ \sqrt[n]{k} } = \lim_{n \to \infty}{ k^{1/n} } \\ ln(y) & = & ln\left[ \lim_{n \to \infty}{ k^{1/n} } \right] \\ & = & \lim_{n \to \infty}{ ln\left( k^{1/n} \right) } \\ & = & \lim_{n \to \infty}{ (1/n)ln(k) } \\ & = & \lim_{n \to \infty}{ \frac{ln(k)}{n} } \\ & = & 0 \\ e^{ln(y)} & = & e^0 \\ y & = & 1 \end{array} }$$

$$\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{k} } = 1 }$$

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Practice B05

$$\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{n^k} } }$$ where k is a positive constant

Practice B06

$$\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{ln(n)} } }$$

Practice B07

$$\displaystyle{ \lim_{x \to 0^+}{\sqrt{x}ln(x)} }$$

Practice B08

$$\displaystyle{ \lim_{x \to 0}{\cot(2x)\sin(6x)} }$$

##### Practice Problem [ B05 ] Solution

$$\displaystyle{ \begin{array}{rcl} y & = & \lim_{n \to \infty}{ \sqrt[n]{n^k} } = \lim_{n \to \infty}{ n^{k/n} } \\ ln(y) & = & ln \left[ \lim_{n \to \infty}{ n^{k/n} } \right] \\ & = & \lim_{n \to \infty}{ ln \left[ n^{k/n} \right] } \\ & = & \lim_{n \to \infty}{ (k/n)ln(n) } \\ & = & \lim_{n \to \infty}{ \frac{kln(n)}{n} } \\ & = & \lim_{n \to \infty}{ \frac{k(1/n)}{1} } \\ & = & \lim_{n \to \infty}{ \frac{k}{n} } \\ & = & 0 \\ e^{ln(y)} & = & e^0 \\ y & = & 1 \end{array} }$$

$$\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{n^k} } = 1 }$$

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##### Practice Problem [ B06 ] Solution

$$\displaystyle{ \begin{array}{rcl} y & = & \lim_{n \to \infty}{ \sqrt[n]{ln(n)} } = \lim_{n \to \infty}{ (ln(n))^{1/n} } \\ ln(y) & = & ln \left[ \lim_{n \to \infty}{(ln(n))^{1/n} } \right] \\ & = & \lim_{n \to \infty}{ ln \left[ (ln(n))^{1/n} \right] } \\ & = & \lim_{n \to \infty}{ (1/n)ln(ln(n)) } \\ & = & \lim_{n \to \infty}{ \frac{ln(ln(n))}{n} } \\ & = & \lim_{n \to \infty}{ \frac{1/n}{ln(n)(1)} } \\ & = & \lim_{n \to \infty}{ \frac{1}{nln(n)} } \\ & = & 0 \\ e^{ln(y)} & = & e^0 \\ y & = & 1 \end{array} }$$

$$\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{ln(n)} } = 1 }$$

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##### Practice Problem [ B07 ] Solution

$$\displaystyle{ \lim_{x \to 0^+}{\sqrt{x}ln(x)} = 0 }$$

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##### Practice Problem [ B08 ] Solution

$$\displaystyle{ \lim_{x \to 0}{\cot(2x)\sin(6x)} = 3 }$$

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Practice B09

$$\displaystyle{ \lim_{x \to \infty}{(xe^{1/x} - x)} }$$

Practice B10

$$\displaystyle{ \lim_{x \to 1}{ \left[ \frac{1}{ln(x)} - \frac{1}{x-1} \right] } }$$

Practice B11

$$\displaystyle{ \lim_{x \to \infty}{(e^x + x) ^{1/x}} }$$

Practice B12

$$\displaystyle{ \lim_{x \to 0}{(cos(3x))^{5/x}} }$$

##### Practice Problem [ B09 ] Solution

$$\displaystyle{ \lim_{x \to \infty}{ ( xe^{1/x} - x ) } = 1 }$$

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##### Practice Problem [ B10 ] Solution

$$\displaystyle{ \lim_{x \to 1}{ \left[ \frac{1}{ln(x)} - \frac{1}{x-1} \right] } = 1/2 }$$

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##### Practice Problem [ B11 ] Solution

$$\displaystyle{ \lim_{x \to \infty}{(e^x + x) ^{1/x}} = e }$$

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##### Practice Problem [ B12 ] Solution

$$\displaystyle{ \lim_{x \to 0}{(cos(3x))^{5/x}} = 1 }$$

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