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17Calculus - L'Hôpital's Rule

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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.

L'Hôpital's Rule is also referred to as L'Hospital's Rule or Bernoulli's Rule.

If you want a complete lecture on this topic, we recommend this video.

Prof Leonard - Evaluating Limits of Indeterminate Forms

video by Prof Leonard

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 \(c\), \(c^-\), \(c^+\), \(\infty\) or \(-\infty\)

Do not use this on determinate forms. It will probably give you the wrong answer.

For details about determinate and indeterminate forms, click here.

This is not the quotient rule. For information about the difference, click here.

Formal Theorem Statement and Proof

Let f and g be functions that are differentiable on an open interval \((a,b)\) containing c, except possibly at c itself. Assume that \(g'(x) \ne 0\) for all x in \((a,b)\), except possibly at c itself. If the limit of \(f(x)/g(x)\) as x approaches c produces the indeterminate form \(0/0\), then
\(\displaystyle{ \lim_{x\to c}{\frac{f(x)}{g(x)}} = \lim_{x \to c}{\frac{f'(x)}{g'(x)}}}\)
provided the limit on the right exists (or is infinite).
This result also applies when the limit of \(f(x)/g(x)\) as x approaches c produces any one of the indeterminate forms \(\infty/\infty\), \((-\infty)/\infty\), \(\infty/(-\infty)\) or \((-\infty)/(-\infty)\).
[source: Larson Calculus]

Here is a video proof of this theorem.

Larson Calculus - Proof - L'Hopital's Rule from Larson Texts [3min-33secs]

video by Larson Calculus

Indeterminate Forms Table

Determinate-Indeterminate Forms Table

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

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. Be careful to check that you have an indeterminate form before each time you want to apply L'Hopitals Rule.
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 quotient rule. Read this next section 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 \(\displaystyle{ \lim_{x \to c}{\frac{n(x)}{d(x)}} = \lim_{x \to c}{\frac{n'(x)}{d'(x)}}}\).

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

Quotient Rule

L'Hôpital's Rule

\(\displaystyle{ f'(x) = \frac{d(x)n'(x) - n(x)d'(x)}{[d(x)]^2} }\)

\(\displaystyle{ \lim_{x \to c}{[f(x)]} = \lim_{x \to c}{\frac{n(x)}{d(x)}} = \lim_{x \to c}{\frac{n'(x)}{d'(x)}} }\)

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 (setting them side-by-side and comparing and contrasting them). 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.

Types of Indeterminate Forms

Notice in the L'Hôpital's Rule equation, we require the limit equation to be in a fraction. This is a strict requirement that cannot be broken. If the limit equation is not a fraction, we cannot use L'Hôpital's Rule. This section contains a discussion of the four main types of equations that you will run across where you need to determine if the limit is an indeterminate form. (This list is found in compact form on the indeterminate forms page. )

Type 1 - - Indeterminate Quotients

Type 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.

Try these practice problems.

Type 1 Practice

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

Basic

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x-\pi/2}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x-\pi/2}\right]}}\)

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.

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

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

Final Answer

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

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\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}=11/4}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}}\)

Solution

First of all, you always want to try direct substitution. When you do this, you get \(0/0\) which is indeterminate.
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.
\(\displaystyle{ \lim_{x \to 2} \left[ \frac{3x^2-x-10}{x^2-4} \right] = }\) \(\displaystyle{ \lim_{x \to 2} \left[ \frac{6x-1}{2x} \right] = }\) \(\displaystyle{ \frac{6(2)-1}{2(2)} = 11/4 }\)
Remember: You can use L'Hôpital's Rule only if you have an indeterminate form.

Final Answer

\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2-x-10}{x^2-4}\right]}=11/4}\)

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\(\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}=108/7}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}}\)

Solution

First of all, you always want to try direct substitution. When you do this, you get \(0/0\) which is indeterminate.
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.

\(\displaystyle{ \lim_{x \to 3}{ \left[\frac{x^4-81}{2x^2-5x-3}\right] } = }\) \(\displaystyle{ \lim_{x \to 3}{ \left[\frac{4x^3}{4x-5}\right] } = \frac{4(3)^3}{4(3)-5} = \frac{108}{7} }\)
Remember: You can use L'Hôpital's Rule only if you have an indeterminate form.

Final Answer

\(\displaystyle{\lim_{x\to3}{\left[\frac{x^4-81}{2x^2-5x-3}\right]}=108/7}\)

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\(\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}=27}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}}\)

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.

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

Note - - We could have used algebra and factoring to solve this problem. See that solution on the finite limits page.

Final Answer

\(\displaystyle{\lim_{x\to27}{\left[\frac{x-27}{x^{1/3}-3}\right]}=27}\)

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\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}=4/3}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}}\)

Solution

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

\(\displaystyle{ \lim_{x \to 1}{\left[ \frac{x^{1/3}-1}{x^{1/4}-1}\right] } }\)

\(\displaystyle{ \lim_{x \to 1}{ \frac{(1/3)x^{-2/3}}{(1/4)x^{-3/4}} } }\)

\(\displaystyle{ \lim_{x \to 1}{ \frac{4}{3} \frac{x^{3/4}}{x^{2/3}} } }\)

\(\displaystyle{ \lim_{x \to 1}{ \frac{4}{3} x^{3/4 - 2/3} } }\)

\(\displaystyle{ \lim_{x \to 1}{ \frac{4}{3} x^{9 / 12 - 8 / 12} } }\)

\(\displaystyle{ \lim_{x \to 1}{ \frac{4}{3} x^{1 / 12} } }\)

\(\displaystyle{ \frac{4}{3} 1^{1 / 12} = \frac{4}{3} }\)

Note - - We could have used algebra and factoring to solve this problem. See that solution on the finite limits page.

Final Answer

\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}-1}{x^{1/4}-1}\right]}=4/3}\)

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\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}=2}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}}\)

Solution

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

\(\begin{array}{rcl} & & \displaystyle{ \lim_{x \to 0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]} } \\ & = & \displaystyle{ \lim_{x \to 0}{\left[\frac{1+\sec^2(x)}{\cos(x)}\right]} } \\ & = & \displaystyle{ \lim_{x \to 0}{\left[\frac{1+[1/\cos^2(x)]}{\cos(x)}\right]} } \\ & = & \displaystyle{ \lim_{x \to 0}{\left[\frac{\cos^2(x)+1}{\cos^3(x)}\right]} } \\ & = & \frac{1+1}{1} = 2 \end{array}\)
After applying L'Hôpital's Rule once, the rest is just trig and algebra.

Final Answer

\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}=2}\)

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{z\to\pi}{\left[\frac{z-\pi}{\sin(z)}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{z\to\pi}{\left[\frac{z-\pi}{\sin(z)}\right]}}\)

Solution

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

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

Final Answer

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

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\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}}\)

Final Answer

\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}=3/4}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}}\)

Solution

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

\(\displaystyle{\lim_{y \to 0}{\left[ \frac{\sin(3y)}{\sin(4y)} \right]} = }\) \(\displaystyle{ \lim_{y \to 0}{\left[ \frac{3\cos(3y)}{4\cos(4y)} \right]} = \frac{3(1)}{4(1)} = 3/4 }\)
Note - - There is another way to work this using the identity \(\displaystyle{ \lim_{x \to 0}{\left[ \frac{\sin(x)}{x} \right]} = 1 }\).

Final Answer

\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}=3/4}\)

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{x+5}{3x+7}\right]}}\)

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

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.

Final Answer

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^3-5x}\right]}}\)

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.
\(\displaystyle{\lim_{x \to \infty}{ \frac{21x^2 + 1}{6x^2} } = }\) \(\displaystyle{\lim_{x \to \infty}{ \frac{42x}{12x} } = \frac{42}{12} = 7/2}\)
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.

Final Answer

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2-3x+12}{x^3+4x+127}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2-3x+12}{x^3+4x+127}\right]}}\)

Solution

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

Final Answer

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to-\infty}{\left[\frac{7x^2+x+21}{11-x}\right]}}\)

Solution

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

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.

Final Answer

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

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\(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{x}}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{x}}}\)

Solution

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\frac{\ln(x)}{x}}}\)

Solution

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\(\displaystyle{\lim_{x\to0^+}{\frac{\ln(x)}{x}}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0^+}{\frac{\ln(x)}{x}}}\)

Solution

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\frac{x}{\ln(1+2e^x)}}}\)

Solution

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\(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}}\)

Final Answer

\(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}=1}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}}\)

Solution

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Final Answer

\(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}=1}\)

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Intermediate

\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}}\)

Final Answer

\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}=-2}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}}\)

Solution

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

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

\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{\sec^2(x)-1}{\cos(x)-1} \right]} }\)

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

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

\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{-2\sin(x) }{\cos^2(x)\sin(x)\cos(x) } \right]} }\)

\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{-2}{\cos^3(x)} \right]} }\)

\(\displaystyle{ \frac{-2}{1} = -2 }\)

Final Answer

\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)-x}{\sin(x)-x}\right]}=-2}\)

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}}\)

Solution

If we try direct substitution first, we get \( \infty / \infty \) which is indeterminate. So we can use L'Hôpital's Rule.
\(\begin{array}{rcl} & & \displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^5}{e^{5x}} \right]} } \\ & = & \displaystyle{ \lim_{x \to \infty}{\left[ \frac{5x^4}{5e^{5x}} \right]} } \\ & = & \displaystyle{ \lim_{x \to \infty}{\left[ \frac{x^4}{e^{5x}} \right]} } \\ & = & \displaystyle{ \lim_{x \to \infty}{\left[ \frac{4x^3}{5e^{5x}} \right]} } \\ & = & \displaystyle{ \lim_{x \to \infty}{\left[ \frac{12x^2}{25e^{5x}} \right]} } \\ & = & \displaystyle{ \lim_{x \to \infty}{\left[ \frac{24x}{125e^{5x}} \right]} } \\ & = & \displaystyle{ \lim_{x \to \infty}{\left[ \frac{24}{625e^{5x}} \right]} } \\ & = & \displaystyle{ \frac{24}{\infty} = 0 } \end{array}\)
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.

Final Answer

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

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Advanced

\(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } }\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } }\)

Hint

Use the substitution \(u=1/x\) and move the exponential to the denominator.

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } }\)

Final Answer

\(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } = 0 }\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } }\)

Hint

Use the substitution \(u=1/x\) and move the exponential to the denominator.

Solution

Direct substitution yields \(0/0\) which is indeterminate. So our next logical step is to use L'Hôpital's Rule. However, when we do that, we get an even more complicated result.
If we use the hint, the limit will approach \(-\infty\) and we have \(\displaystyle{ \lim_{u \to -\infty}{ \frac{-e^{u}}{1/u} } = \lim_{u \to -\infty}{ \frac{-u}{e^{-u}} } }\).
Now we can use L'Hôpital's Rule (since direct substitution yields the indeterminate fraction \(\infty / \infty\)) to get
\(\displaystyle{ \lim_{u \to -\infty}{ \frac{-u}{e^{-u}} } = }\) \(\displaystyle{ \lim_{u \to -\infty}{ \frac{-1}{-e^{-u}} } = }\) \(\displaystyle{ \frac{1}{e^{\infty}} = 0 }\)

Final Answer

\(\displaystyle{ \lim_{x \to 0^-}{ \frac{-e^{1/x}}{x} } = 0 }\)

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Type 2 - - Indeterminate Products

Type 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.

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Type 2 Practice

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

\(\displaystyle{\lim_{x\to0^+}{\sqrt{x}\ln(x)}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0^+}{\sqrt{x}\ln(x)}}\)

Solution

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\(\displaystyle{\lim_{x\to0}{\cot(2x)\sin(6x)}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\cot(2x)\sin(6x)}}\)

Solution

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Type 3 - - Indeterminate Differences

Type 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 type 1.

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Type 3 Practice

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{(x^2-x)}}\)

Solution

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{(xe^{1/x}-x)}}\)

Solution

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\(\displaystyle{\lim_{x\to1}{\left[\frac{1}{\ln(x)}-\frac{1}{x-1}\right]}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to1}{\left[\frac{1}{\ln(x)}-\frac{1}{x-1}\right]}}\)

Solution

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Type 4 - - Indeterminate Powers

Type 4 - - Indeterminate Powers
In the previous three types, you either started with a fraction in the right form (type 1) or did some algebra to get your function into the right form (type 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 types 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 \(e^L\).

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Type 4 Practice

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

\(\displaystyle{\lim_{n\to\infty}{n^{2/n}}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{n^{2/n}}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{n^{2/n}}}\)

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
\(\begin{array}{rcl} \ln(y) & = & \displaystyle{ \lim_{n \to \infty}{\ln(n^{2/n})} } \\ & = & \displaystyle{ \lim_{n \to \infty}{\frac{2\ln(n)}{n}} } \\ & = & \displaystyle{ \lim_{n \to \infty}{\frac{2/n}{1}} = 0 } \\ \ln(y) & = & 0 \\ e^{\ln(y)} & = & e^0 \\ y & = & 1 \end{array}\)

Final Answer

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}}\) where \(k\) is a positive constant

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}}\) where \(k\) is a positive constant

Solution

\(\displaystyle{ y = \lim_{n \to \infty}{ \sqrt[n]{k} } = \lim_{n \to \infty}{ k^{1/n} } }\)

\(\displaystyle{ \ln(y) = \ln\left[ \lim_{n \to \infty}{ k^{1/n} } \right] }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \ln\left( k^{1/n} \right) } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ (1/n)\ln(k) } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{\ln(k)}{n} } }\)

\( \ln(y) = 0 \)

\( e^{\ln(y)} = e^0 \)

\( y = 1 \)

Final Answer

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}}\) where \(k\) is a positive constant

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}}\) where \(k\) is a positive constant

Solution

\(\displaystyle{ y = \lim_{n \to \infty}{ \sqrt[n]{n^k} } = \lim_{n \to \infty}{ n^{k/n} } }\)

\(\displaystyle{ \ln(y) = \ln \left[ \lim_{n \to \infty}{ n^{k/n} } \right] }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \ln \left[ n^{k/n} \right] } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ (k/n)\ln(n) } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{k\ln(n)}{n} } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{k(1/n)}{1} } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{k}{n} } }\)

\( \ln(y) = 0 \)

\( e^{\ln(y)} = e^0 \)

\( y = 1 \)

Final Answer

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}}\)

Final Answer

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}}\)

Solution

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

\(\displaystyle{ \ln(y) = \ln \left[ \lim_{n \to \infty}{(\ln(n))^{1/n} } \right] }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \ln \left[ (\ln(n))^{1/n} \right] } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ (1/n)\ln(\ln(n)) } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{\ln(\ln(n))}{n} } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{1/n}{\ln(n)(1)} } }\)

\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{1}{n\ln(n)} } }\)

\( e^{\ln(y)} = e^0 \)

\(y = 1\)

Final Answer

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

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

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to\infty}{(e^x+x)^{1/x}}}\)

Solution

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\(\displaystyle{\lim_{x\to0}{(\cos(3x))^{5/x}}}\)

Problem Statement

Evaluate this limit using L'Hopital's Rule (if valid), giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{(\cos(3x))^{5/x}}}\)

Solution

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\(\displaystyle{ \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } }\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{ \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } }\).

Hint

Use natural logarithms to get the \(x\) out of the exponent.

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{ \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } }\).

Final Answer

\(\displaystyle{ \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } = \frac{1}{e} }\)

Problem Statement

Evaluate this limit, giving your answer in exact terms. \(\displaystyle{ \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } }\).

Hint

Use natural logarithms to get the \(x\) out of the exponent.

Solution

First let's assign the limit to the variable \(y\).

\(\displaystyle{ y = \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } }\)

Use the hint by taking the natural logarithm of both sides of the equation.

\(\displaystyle{ \ln(y) = \ln \left[ \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } \right] }\)

On the left, move the natural logarithm inside of the limit. We can do this since the natural log function is continuous.

\(\displaystyle{ \ln(y) = \lim_{x \to \infty}{ \ln \left( \frac{x}{x+1} \right)^x } }\)

Now we can move the exponent out in front using the basic logarithm law \(\ln(x^y) = y\ln x\)

\(\displaystyle{ \ln(y) = \lim_{x \to \infty}{ x \ln \left( \frac{x}{x+1} \right) } }\)

In order to use L'Hopital's Rule, we need \(0/0\) or \(\infty / \infty\). Let's move the \(x\) out in front to the denominator.

\(\displaystyle{ \ln(y) = \lim_{x \to \infty}{ \frac{ \ln (x/(x+1)) }{1/x} } }\)

Now we have a limit that goes to \(0/0\). (If you are not sure about the top going to zero, convince yourself by working it on the side. One way is to multiply the numerator and denominator by \(1/x\) and taking the limit of numerator and denominator separately.) So now we can apply L'Hopital's Rule.

Let's take the derivatives of the numerator and denominator individually. First the numerator.

\(\displaystyle{ \frac{d}{dx} \ln (x/(x+1)) }\)

Let's break the natural logarithm into two terms to make the derivative easier to evaluate.

\(\displaystyle{ \frac{d}{dx} [ \ln(x) - \ln(x+1) ] }\)

Now take the derivative.

\(\displaystyle{ \frac{1}{x} - \frac{1}{x+1} = \frac{x+1-x}{x(x+1)} = \frac{1}{x(x+1)} }\)

Now evaluate the derivative of the denominator.

\(\displaystyle{ \frac{d[1/x]}{dx} = \frac{d[x^{-1}]}{dx} = -x^{-2} }\)

Now let's put the results of the derivatives back into the limit to see what we have.

\(\displaystyle{ \ln(y) = \lim_{x \to \infty}{ \frac{1}{x(x+1)} \left[ \frac{1}{-x^{-2}} \right] } }\)

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

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

\(\displaystyle{ \ln(y) = \lim_{x \to \infty}{ \frac{-x}{x+1} } }\)

\(\displaystyle{ \ln(y) = \lim_{x \to \infty}{ \frac{-x(1/x)}{(x+1)(1/x)} } }\)

\(\displaystyle{ \ln(y) = \lim_{x \to \infty}{ \frac{-1}{1+1/x} } }\)

Taking the limit gives us \(-1/(1+0) = -1\)

So our equation is now \( \ln(y) = -1 \)

Now we need to solve for \(y\) which is the original limit.

\( e^{\ln y} = e^{-1} \to y = 1/e\)

So, a question that you may have is, why did we assign the original limit to the variable \(y\) when we didn't really use it except at the very end of the problem? Couldn't we have just left it off? Yes, we could have done without it. However, we have found that it is very easy to forget to convert back at the end of the problem and just say the limit is \(-1\) in this case. So we use it as a placeholder and as a reminder that we have that last step before we get our final answer. Notice that we carry it along in the problem, so that we don't forget it. This is important too.
Of course, check with your instructor to see what they expect.

Final Answer

\(\displaystyle{ \lim_{x \to \infty}{ \left( \frac{x}{x+1} \right)^x } = \frac{1}{e} }\)

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How It Works

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 a video to help explain how it works and see some examples. This 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.

MIT OCW - Lec 35 | MIT 18.01 Single Variable Calculus, Fall 2007 [48min-42secs]

video by MIT OCW

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Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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