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L'Hôpital's Rule 
L'Hôpital's Rule is also referred to as L'Hospital's Rule or Bernoulli's Rule. 
on this page:
► theorem ► some things to notice ► types of indeterminate forms ► indeterminate forms table

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

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  LHopitals Rule Proof [3min33secs] 


Quick Notes
This rule can be repeated until a determinate form is found. 
\(x\) can approach a finite value \(c\), \(c^\), \(c^+\), \(\infty\) or \(\infty\) 
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. 

Indeterminate Forms Table
DeterminateIndeterminate 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 
download this table


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 this next section to clarify it in your mind.
Difference Between L'Hôpital's Rule and the Quotient Rule
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 
\(\displaystyle{ f'(x) = \frac{d(x)n'(x)  n(x)d'(x)}{[d(x)]^2} }\)

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

If you haven't already, now would be a good time to study what is meant by an indeterminate form, since the following discussion requires you to know it. You can find the complete discussion on the indeterminate forms page.
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 also found on the indeterminate forms page. )
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 13 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 \(e^L\).

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 indepth 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 

Instructions   Evaluate the following limits using L'Hôpital's Rule ( if valid ), giving your answers in exact form.
Practice A01 
\(\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x\pi/2}\right]}}\) 


\(\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x\pi/2}\right]}=2}\) 
If we plug in \( \pi/2 \) for x, we get \( 0/0 \) which is indeterminate. This means we can use L'HÃ´pital'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).
Practice A01 Final Answer 
\(\displaystyle{\lim_{x\to\pi/2}{\left[\frac{\tan(2x)}{x\pi/2}\right]}=2}\) 
Practice A02 
\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2x10}{x^24}\right]}}\) 


\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2x10}{x^24}\right]}=11/4}\) 
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.
\(\displaystyle{ \lim_{x \to 2} \left[ \frac{3x^2x10}{x^24} \right] = }\) \(\displaystyle{
\lim_{x \to 2} \left[ \frac{6x1}{2x} \right] = }\) \(\displaystyle{
\frac{6(2)1}{2(2)} = 11/4 }\)
Remember: You can use L'HÃ´pital's Rule only if you have an indeterminate form.
Practice A02 Final Answer 
\(\displaystyle{\lim_{x\to2}{\left[\frac{3x^2x10}{x^24}\right]}=11/4}\) 
Practice A03 
\(\displaystyle{\lim_{x\to3}{\left[\frac{x^481}{2x^25x3}\right]}}\) 


\(\displaystyle{\lim_{x\to3}{\left[\frac{x^481}{2x^25x3}\right]}=108/7}\) 
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.
\(\displaystyle{
\lim_{x \to 3}{ \left[\frac{x^481}{2x^25x3}\right] } = }\) \(\displaystyle{
\lim_{x \to 3}{ \left[\frac{4x^3}{4x5}\right] } = \frac{4(3)^3}{4(3)5} = 108/7 }\)
Remember: You can use L'Hôpital's Rule only if you have an indeterminate form.
Practice A03 Final Answer 
\(\displaystyle{\lim_{x\to3}{\left[\frac{x^481}{2x^25x3}\right]}=108/7}\) 
Practice A04 
\(\displaystyle{\lim_{x\to27}{\left[\frac{x27}{x^{1/3}3}\right]}}\) 


\(\displaystyle{\lim_{x\to27}{\left[\frac{x27}{x^{1/3}3}\right]}=27}\) 
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.
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to 27}{\left[ \frac{x  27}{x^{1/3}  3}\right]} \\\\
& = & \lim_{x \to 27}{ \frac{1}{(1/3)x^{2/3}} } \\\\
& = & \lim_{x \to 27}{ \frac{x^{2/3}}{1/3} } \\\\
& = & \lim_{x \to 27}{ 3x^{2/3} } \\\\
& = & 3(27)^{2/3} = 3(9) = 27
\end{array}
}\)
Note   We could have used algebra and factoring to solve this problem. See that solution on the finite limits page.
Practice A04 Final Answer 
\(\displaystyle{\lim_{x\to27}{\left[\frac{x27}{x^{1/3}3}\right]}=27}\) 
Practice A05 
\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}1}{x^{1/4}1}\right]}}\) 


\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}1}{x^{1/4}1}\right]}=4/3}\) 
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.
Practice A05 Final Answer 
\(\displaystyle{\lim_{x\to1}{\left[\frac{x^{1/3}1}{x^{1/4}1}\right]}=4/3}\) 
Practice A06 
\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}}\) 


\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}=2}\) 
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.
Practice A06 Final Answer 
\(\displaystyle{\lim_{x\to0}{\left[\frac{x+\tan(x)}{\sin(x)}\right]}=2}\) 
Practice A07 
\(\displaystyle{\lim_{z\to\pi}{\left[\frac{z\pi}{\sin(z)}\right]}}\) 


\(\displaystyle{\lim_{z\to\pi}{\left[\frac{z\pi}{sin(z)}\right]}=1}\) 
Direct substitution yields \(0/0\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{z \to \pi}{\left[ \frac{z  \pi}{sin(z)} \right]} = }\) \(\displaystyle{
\lim_{z \to \pi}{\left[ \frac{1}{cos(z)} \right]} = }\) \(\displaystyle{
\frac{1}{1} = 1 }\)
Practice A07 Final Answer 
\(\displaystyle{\lim_{z\to\pi}{\left[\frac{z\pi}{sin(z)}\right]}=1}\) 
Practice A08 
\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}}\) 


\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}=3/4}\) 
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 \(\displaystyle{ \lim_{x \to 0}{\left[ \frac{\sin(x)}{x} \right]} = 1 }\).
Practice A08 Final Answer 
\(\displaystyle{\lim_{y\to0}{\left[\frac{\sin(3y)}{\sin(4y)}\right]}=3/4}\) 
Practice A09 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x+5}{3x+7}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\) 
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.
Practice A09 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x+5}{3x+7}\right]}=1/3}\) 
Practice A10 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^35x}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^35x}\right]}=7/2}\) 
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.
Practice A10 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^3+x+12}{2x^35x}\right]}=7/2}\) 
Practice A11 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^23x+12}{x^3+4x+127}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^23x+12}{x^3+4x+127}\right]}=0}\) 
Direct substitution yields \(\displaystyle{ \frac{\infty  \infty + 12}{\infty + \infty + 127} }\) which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{x \to \infty}{\left[ \frac{7x^2  3x + 12}{x^3 + 4x + 127} \right]} = }\) \(\displaystyle{
\lim_{x \to \infty}{\left[ \frac{14x  3}{3x^2 + 4} \right]}}\)
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.
Practice A11 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^23x+12}{x^3+4x+127}\right]}=0}\) 
Practice A12 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2+x+21}{11x}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2+x+21}{11x}\right]}=+\infty}\) 
Direct substitution yields \(\infty / \infty\) which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{\lim_{x \to \infty}{\left[ \frac{7x^2 + x + 21}{11x} \right]} = }\) \(\displaystyle{
\lim_{x \to \infty}{ \frac{14x + 1}{1} }}\)
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.
Practice A12 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{7x^2+x+21}{11x}\right]}=+\infty}\) 
Practice A13 
\(\displaystyle{\lim_{x\to0}{\frac{\sin(x)}{x}}}\) 

Practice A14 
\(\displaystyle{\lim_{x\to\infty}{\frac{\ln(x)}{x}}}\) 

Practice A15 
\(\displaystyle{\lim_{x\to0^+}{\frac{\ln(x)}{x}}}\) 

Practice A16 
\(\displaystyle{\lim_{x\to\infty}{\frac{x}{\ln(1+2e^x)}}}\) 

Practice A17 
\(\displaystyle{\lim_{x\to\infty}{(x^2x)}}\) 

Practice A18 
\(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}}\) 


\(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}=1}\) 
Practice A18 Final Answer 
\(\displaystyle{\lim_{x\to 0}{\frac{\sin(x)}{x+x^2}}=1}\) 
Practice B01 
\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)x}{\sin(x)x}\right]}}\) 


\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)x}{\sin(x)x}\right]}=2}\) 
Direct substitution yields \(0/0\), which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{\tan(x)x}{\sin(x)x} \right]} }\) 
\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{\sec^2(x)1}{\cos(x)1} \right]} }\) 
\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{2\sec(x)[\sec(x)\tan(x)]}{\sin(x)} \right]} }\) 
\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{2\sec^2(x)\tan(x)}{\sin(x)} \right]} }\) 
\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{2\sin(x) }{\cos^2(x)\sin(x)\cos(x) } \right]} }\) 
\(\displaystyle{ \lim_{x \to 0}{\left[ \frac{2}{\cos^3(x)} \right]} }\) 
\(\displaystyle{ \frac{2}{1} = 2 }\) 
Practice B01 Final Answer 
\(\displaystyle{\lim_{x\to0}{\left[\frac{\tan(x)x}{\sin(x)x}\right]}=2}\) 
Practice B02 
\(\displaystyle{\lim_{n\to\infty}{n^{2/n}}}\) 


\(\displaystyle{\lim_{n\to\infty}{n^{2/n}}=1}\) 
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
\(\displaystyle{
\begin{array}{rcl}
\ln(y) & = & \lim_{n \to \infty}{\ln(n^{2/n})} \\\\
& = & \lim_{n \to \infty}{\frac{2\ln(n)}{n}} \\\\
& = & \lim_{n \to \infty}{\frac{2/n}{1}} = 0 \\\\
\ln(y) & = & 0 \\\\
e^{\ln(y)} & = & e^0 \\\\
y & = & 1
\end{array}
}\)
Practice B02 Final Answer 
\(\displaystyle{\lim_{n\to\infty}{n^{2/n}}=1}\) 
Practice B03 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}}\) 


\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}=0}\) 
If we try direct substitution first, we get \( \infty/\infty \) which is indeterminate. So we can use L'Hôpital's Rule.
\(\displaystyle{
\begin{array}{rcl}
& & \lim_{x \to \infty}{\left[ \frac{x^5}{e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{5x^4}{5e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{x^4}{e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{4x^3}{5e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{12x^2}{25e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{24x}{125e^{5x}} \right]} \\\\
& = & \lim_{x \to \infty}{\left[ \frac{24}{625e^{5x}} \right]} \\\\
& = & \frac{24}{\infty} = 0
\end{array}}\)
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.
Practice B03 Final Answer 
\(\displaystyle{\lim_{x\to\infty}{\left[\frac{x^5}{e^{5x}}\right]}=0}\) 
Practice B04 
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}}\) where k is a positive constant 


\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}=1}\) 
\(\displaystyle{ y = \lim_{n \to \infty}{ \sqrt[n]{k} } = \lim_{n \to \infty}{ k^{1/n} } }\) 
\(\displaystyle{ \ln(y) = \ln\left[ \lim_{n \to \infty}{ k^{1/n} } \right] }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \ln\left( k^{1/n} \right) } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ (1/n)\ln(k) } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{\ln(k)}{n} } }\) 
\(\displaystyle{ \ln(y) = 0 }\) 
\(\displaystyle{ e^{\ln(y)} = e^0 }\) 
\(\displaystyle{ y = 1 }\) 
Practice B04 Final Answer 
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{k}}=1}\) 
Practice B05 
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}}\) where k is a positive constant 


\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}=1}\) 
\(\displaystyle{ y = \lim_{n \to \infty}{ \sqrt[n]{n^k} } = \lim_{n \to \infty}{ n^{k/n} } }\) 
\(\displaystyle{ \ln(y) = \ln \left[ \lim_{n \to \infty}{ n^{k/n} } \right] }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \ln \left[ n^{k/n} \right] } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ (k/n)\ln(n) } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{k\ln(n)}{n} } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{k(1/n)}{1} } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{k}{n} } }\) 
\(\displaystyle{ \ln(y) = 0 }\) 
\(\displaystyle{ e^{\ln(y)} = e^0 }\) 
\( y = 1 \) 
Practice B05 Final Answer 
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{n^k}}=1}\) 
Practice B06 
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}}\) 


\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}=1}\) 
\(\displaystyle{ y = \lim_{n \to \infty}{ \sqrt[n]{\ln(n)} } = \lim_{n \to \infty}{ (\ln(n))^{1/n} } }\) 
\(\displaystyle{ \ln(y) = \ln \left[ \lim_{n \to \infty}{(\ln(n))^{1/n} } \right] }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \ln \left[ (\ln(n))^{1/n} \right] } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ (1/n)\ln(\ln(n)) } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{\ln(\ln(n))}{n} } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{1/n}{\ln(n)(1)} } }\) 
\(\displaystyle{ \ln(y) = \lim_{n \to \infty}{ \frac{1}{n\ln(n)} } }\) 
\(\displaystyle{ e^{\ln(y)} = e^0 }\) 
\(y = 1\) 
Practice B06 Final Answer 
\(\displaystyle{\lim_{n\to\infty}{\sqrt[n]{\ln(n)}}=1}\) 
Practice B07 
\(\displaystyle{\lim_{x\to0^+}{\sqrt{x}\ln(x)}}\) 

Practice B08 
\(\displaystyle{\lim_{x\to0}{\cot(2x)\sin(6x)}}\) 

Practice B09 
\(\displaystyle{\lim_{x\to\infty}{(xe^{1/x}x)}}\) 

Practice B10 
\(\displaystyle{\lim_{x\to1}{\left[\frac{1}{\ln(x)}\frac{1}{x1}\right]}}\) 

Practice B11 
\(\displaystyle{\lim_{x\to\infty}{(e^x+x)^{1/x}}}\) 

Practice B12 
\(\displaystyle{\lim_{x\to0}{(\cos(3x))^{5/x}}}\) 

Practice C01 
\(\displaystyle{ \lim_{x \to 0^}{ \frac{e^{1/x}}{x} } }\) 


\(\displaystyle{ \lim_{x \to 0^}{ \frac{e^{1/x}}{x} } = 0 }\) 
Direct substitution yields \(0/0\) which is indeterminate. So our next logical step is to use L'HÃ´pital'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'HÃ´pital's Rule (since direct substitution yields the indeterminate fraction \(\infty/\infty\)) to get
\(\displaystyle{ \lim_{u \to \infty}{ \frac{u}{e^{u}} } = }\)
\(\displaystyle{ \lim_{u \to \infty}{ \frac{1}{e^{u}} } = }\)
\(\displaystyle{ \frac{1}{e^{\infty}} = 0 }\)
Practice C01 Final Answer 
\(\displaystyle{ \lim_{x \to 0^}{ \frac{e^{1/x}}{x} } = 0 }\) 