L'Hopital's Rule

In introductory calculus we learned that we may use differentiation to sketch the graphs of functions.

Practically, we may instead plot graphs numerically with/without the aid of computers.

Find out the value of the desired function at a point by calculation or by using graphs of simpler functions, then repeat for as many points as you like.

Examples.

1. $y=\sin x+\cos x$.



2. $y=x \sin x$.

The following functions are undefined at $x=0$.

Therefore, we cannot find their limits at $x=0$ using "direct substitution".

However, we may still guess the limits by looking at their graphs.

1. $\displaystyle \lim\limits_{x\to 0}x \ln x=$ $0$ 1 1/2 0 -1/2 -1 Oops! Please try again



2. $\displaystyle \lim\limits_{x\to 0}\frac{\ln \sin 2x}{\ln \sin x}=$ $1$ 1 1/2 0 -1/2 -1 Oops! Please try again



3. $\displaystyle \lim\limits_{x\to 0}\frac{\ln (1+x)-x\cos x}{x^2}=$ $\frac{-1}{2}$ 1 1/2 0 -1/2 -1 Oops! Please try again

We learned two limit theorems:

Why are they true?

Hint:

• In both cases, both the denominator and the numerator approach zero.
• Try to comment on the similarity between the denominator and the numerator near zero.
• In other words, observe and compare how the denominator and the numerator approach zero, in particular, their direction and speed.

Let $h(x)=\frac{f(x)}{g(x)}$.

Intuitively, we want to write $\lim\limits_{x\to a}h(x)=\frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)}$. $\quad \quad \quad \textbf{(1)}$

However, if $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)=0, \textbf{(1)}$ clearly does not hold.

Not only is the expression $\frac{0}{0}$ undefined, we are also unable to make any conclusion on $\lim\limits_{x\to a}h(x)$.

The following demonstrates two examples where $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)=0$ in both examples but $\lim\limits_{x\to a}h(x)$ are clearly different.

1. $f(x)=\sin x, g(x)=x, x\to 0, \lim\limits_{x\to 0}\frac{f(x)}{g(x)}=1$.
2. $f(x)=-\sin x, g(x)=x, x\to 0, \lim\limits_{x\to 0}\frac{f(x)}{g(x)}=-1$.

When $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)=0$, the limit $\lim\limits_{x\to a}\frac{f(x)}{g(x)}$ is said to lead to $\frac{0}{0}$ (an indeterminate form).

Other indeterminate forms are $\dfrac{\infty}{\infty},~0 \times \infty,~\infty - \infty,~0^0,~1^{\infty}~{\rm{and}}~\infty^{0}$.

In this part, we shall explore how different indeterminate forms can be converted into $\frac{0}{0}$ form.

1. $\frac{\infty}{\infty}, \quad 0\times \infty, \quad \infty - \infty$ forms.

   A. $ \quad \frac{f(x)}{\frac{1}{g(x)}}$ B. $ \quad \frac{\frac{1}{g(x)}-\frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}$ C. $ \quad \frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}$

   indeterminate form (before)	$\lim\limits_{x\to a}f(x)$	$\lim\limits_{x\to a}g(x)$	target limit $\lim\limits_{x\to a}h(x)$	transformation	indeterminate form (after)
   $\frac{\infty}{\infty}$	$\infty$	$\infty$	$h(x)=\frac{f(x)}{g(x)}$	$h(x)=$ A B C Oops! Please try again $\frac{\frac{1}{g(x)}-\frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}$	$\frac{0}{0}$
   $0\times \infty$	$0$	$\infty$	$h(x)=f(x)g(x)$	$h(x)=$ A B C Oops! Please try again $\frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}$	$\frac{0}{0}$
   $\infty - \infty$	$\infty$	$\infty$	$h(x)=f(x)-g(x)$	$h(x)=$ A B C Oops! Please try again $\frac{f(x)}{\frac{1}{g(x)}}$	$\frac{0}{0}$





2. $0^0, \quad 1^\infty, \quad \infty^0$ forms.

   Note that if $\displaystyle \lim_{x\to a} \ln h(x)=L$, then $\displaystyle \lim_{x\to a} h(x)=e^L$.

   indeterminate form (before)	$\lim\limits_{x\to a}f(x)$	$\lim\limits_{x\to a}g(x)$	target limit $\lim\limits_{x\to a}h(x)$	transformation	indeterminate form (after)
   $0^0$	$0$	$0$	$h(x)=f(x)^{g(x)}$	$\ln h(x)=g(x) \ln f(x)$	$0 \times \infty$
   $1^\infty$	$1$	$\infty$	$h(x)=f(x)^{g(x)}$	$\ln h(x)=g(x) \ln f(x)$	$\infty \times 0$
   $\infty^0$	$\infty$	$0$	$h(x)=f(x)^{g(x)}$	$\ln h(x)=g(x) \ln f(x)$	$0 \times \infty$

Exercises.

Determine whether the following limits are indeterminate. For those that are indeterminate, identify their indeterminate forms. (work out how to convert them into $\frac{0}{0}$ form by yourself!)

1. $\lim\limits_{x\to 0}x^x$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
2. $\lim\limits_{x\to 0}\frac{5\sin x-7 \sin 2x+3\sin 3x}{\tan x-x}$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
3. $\lim\limits_{x\to 1}(1-x)\tan(\frac{\pi x}{2})$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
4. $\lim\limits_{x\to 0}(\frac{1}{x^2}-\csc^2 x)$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
5. $\lim\limits_{x\to 0}\frac{\sin x - \cos x}{\sin x + \cos x}$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
6. $\lim\limits_{x\to 0}\frac{\ln x}{\cot x}$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
7. $\lim\limits_{x\to 0}\frac{\ln(1+x^2)}{1-\cos x}$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
8. $\lim\limits_{x\to 0}\frac{e^x+\ln (1-x)-1}{\tan x-x}$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!
9. $\lim\limits_{x\to \frac{\pi}{2}}(\tan x)^{\cos x}$
   $\frac{0}{0}$ $\frac{\infty}{\infty}$ $0\times \infty$ $\infty - \infty$ $0^0$ $1^\infty$ $\infty^0$ not indeterminate Oops! Please try again Correct!

In this part, we shall explore how Taylor series expansions lead to the result of L'Hopital's Rule. (For a proof, visit part 9)

Examples.

Expand $\sin x, \quad \cos x, \quad \sin ^2 x, \quad \cos ^2 x, \quad \ln (1+x)$ and $e^x$ as Taylor series around $x=0$. Hence, find the following limits.

1. $\lim\limits_{x\to 0} \frac{x-\sin x}{x^3}$.
2. $\lim\limits_{x\to 0}\frac{x \cos x-\ln(1+x)}{x^2}$.
3. $\lim\limits_{x\to 0} \frac{\cos x-\ln(1+x)+\sin x -1}{e^x-(1+x)}$.
4. $\lim\limits_{x\to 0} (\frac{1}{x^2}-\cot^2 x)$.
5. $\lim\limits_{x\to 0} \frac{x^{1/2}\tan x}{(e^x-1)^{3/2}}$.

Solution.

Taylor series expansions:

1. $x-\frac{x^2}{2}+\frac{x^3}{6}+\cdots$
2. $1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$
3. $1-x^2+\frac{x^4}{3}-\frac{2x^6}{45}+\cdots$
4. $x^2-\frac{x^4}{3}+\frac{2x^6}{45}-\cdots$
5. $x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots$
6. $1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots$

In this part, we will try to visualize the meaning of L'Hopital's Rule.

L'Hopital's Rule ($\frac{0}{0}$ form) states that if $\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)=0$, then $\lim\limits_{x\to a}\frac{f(x)}{g(x)}=\lim\limits_{x\to a}\frac{f'(x)}{g'(x)}$.

Examples.

1. $f(x)=\sin x, g(x)=x, a=0$. Then $\lim\limits_{x\to 0}\frac{\sin x}{x}=\lim\limits_{x\to 0}\frac{\cos x}{1}=1$.
2. $f(x)=e^x-1, g(x)=x, a=0$. Then $\lim\limits_{x\to 0}\frac{e^x-1}{x}=\lim\limits_{x\to 0}\frac{e^x}{1}=1$.

In this part, we shall see how L'Hopital's Rule can be used to evaluate limits.

Exercises.

Find the values of the following limits. (Try to solve them by yourself before reading the solution! Hint: see part 5)

1. $\lim\limits_{x\to 0}\frac{x \cos x-\ln(1+x)}{x^2}$$=\frac{1}{2}$see numerical answer
   view complete solution

   Solution.

   \begin{align} & \lim\limits_{x\rightarrow 0} \dfrac{ x \cos x- \ln (1 + x)}{ x^2} & & \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x \rightarrow 0} \dfrac{ \dfrac{d\left( x \cos x- \ln (1 + x)\right)}{dx}}{ \dfrac{ d \left( x^2 \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \cos x - x \sin x - \dfrac{1}{1+ x}}{ 2x } & & \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d\left( \cos x - x \sin x - \dfrac{1}{1+ x} \right)}{dx}}{ \dfrac{ d \left( 2 x \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ -\sin x - \sin x - x \cos x + \dfrac{1}{(1+ x)^2} }{ 2} & & \\ = & \dfrac{ 1}{ 2 }. \end{align}


2. $\lim\limits_{x\to 0} \frac{\cos x-\ln(1+x)+\sin x -1}{e^x-(1+x)}$$=0$see numerical answer
   view complete solution

   Solution.

   \begin{align} & \lim\limits_{x\rightarrow 0} \dfrac{ \cos x - \ln (1+ x) + \sin x - 1}{ e^x - (1 + x)} & & \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( \cos x - \ln (1+ x) + \sin x - 1)}{dx}}{ \dfrac{ d \left( e^x - (1 + x) \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ - \sin x + \dfrac{1}{x+1} - \cos x}{ e^x - 1} & & \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d\left( - \sin x + \dfrac{1}{x+1} - \cos x \right)}{dx}}{ \dfrac{ d \left( e^x - 1 \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ - \cos x + \dfrac{1}{(1+x)^2} - \sin x}{ e^x} & & \\ = & 0. \end{align}


3. $\lim\limits_{x\to 0} \frac{\ln x}{\cot x}$$=0$see numerical answer
   view complete solution

   Solution.

   \begin{align} & \lim\limits_{x\rightarrow 0} \dfrac{ \ln x }{ \cot x} & & \dfrac{\infty}{\infty}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( \ln x )}{dx}}{ \dfrac{ d \left( \cot x \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{1}{x} }{ - {\mathrm{cosec}}^2 x} \\ = & - \lim\limits_{x\rightarrow 0} \dfrac{ \sin^2 x}{ x} & & \dfrac{0}{0}~{\rm {form}} \\ = & - \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( \sin^2 x )}{dx}}{ \dfrac{ d \left( x \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & - \lim\limits_{x\rightarrow 0} \dfrac{ 2 \sin x \cos x }{ 1 } & & \\ = & 0. \end{align}


4. $\lim\limits_{x\to 1} (1-x)\tan(\frac{\pi x}{2})$$=\frac{2}{\pi}$see numerical answer
   view complete solution

   Solution.

   \begin{align} & \lim\limits_{x\rightarrow 0} (1 - x) \tan \left( \dfrac{ \pi x }{ 2} \right) & & 0 \times \infty ~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ 1 - x }{ \cot \left( \dfrac{ \pi x }{ 2} \right) } & & {\text{transform to }} \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( 1 - x )}{dx}}{ \dfrac{ d \left( \cot \left( \dfrac{ \pi x }{ 2} \right) \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ -1 }{ - \dfrac{ \pi }{ 2 } {\mathrm{cosec}}^2 \left( \dfrac{ \pi x }{ 2} \right) } & & \\ = & \dfrac{ 2 }{ \pi }. \end{align}


5. $\lim\limits_{x\to 0}(\frac{1}{x^2}-\csc^2 x)$$=\frac{-1}{3}$see numerical answer
   view complete solution

   Solution.

   \begin{align} & \lim\limits_{x\rightarrow 0} \left( \dfrac{1}{x^2} - \csc^2 x \right) & & \infty - \infty ~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \sin^2 x - x^2 }{ x^2 \sin^2 x } & & \text{tranform to } \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( \sin^2 x - x^2 )}{dx}}{ \dfrac{ d \left( x^2 \sin^2 x \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \sin 2x - 2x }{ x^2 \sin 2x + 2x \sin^2 x } & & \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( \sin 2x - 2x )}{dx}}{ \dfrac{ d \left( x^2 \sin 2x + 2x \sin^2 x \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ 2 \cos 2x - 2 }{ 2 x^2 \cos 2x + 4x \sin 2x + 2 \sin^2 x } & & \dfrac{0}{0} ~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( 2 \cos 2x - 2 )}{dx}}{ \dfrac{ d \left( 2 x^2 \cos 2x + 4x \sin 2x + 2 \sin^2 x \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ -4 \sin 2x }{ - 4 x^2 \sin 2x + 12 x \cos 2x + 6 \sin 2x } & & \dfrac{0}{0}~{\rm {form}} \\ = & \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( -4 \sin 2x )}{dx}}{ \dfrac{ d \left( - 4 x^2 \sin 2x + 12 x \cos 2x + 6 \sin 2x \right)}{dx}} & & {\text {L'Hopital's Rule}}\\ = & \lim\limits_{x\rightarrow 0} \dfrac{ 8 \cos 2x }{ -8 x^2 \cos 2x - 32 \sin 2x + 24 \cos 2x } & & \\ = & - \dfrac{ 8 }{ 24 } = - \dfrac{1}{ 3 }. \end{align}


6. $\lim\limits_{x\to 0}x^x$$=1$see numerical answer
   view complete solution

   Solution.

   Let $f(x)=\lim\limits_{x\to 0} x^x$ $(0^0 \text{ form})$.

   \begin{align} \ln f(x) &= \lim\limits_{x\rightarrow 0} x \ln x & & \text{transform to } 0 \times \infty ~{\rm {form}} \\ & = \lim\limits_{x\rightarrow 0} \frac{ \ln x}{ \dfrac{1}{x}} & & \text{transform to } \dfrac{\infty}{\infty}~{\rm {form}} \\ & = \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{d( \ln x )}{dx}}{ \dfrac{ d \left( \dfrac{1}{x} \right)}{dx}} & & {\text {L'Hopital's Rule}} \\ & = \lim\limits_{x\rightarrow 0} \dfrac{ \dfrac{1}{x} }{ \dfrac{-1}{x^2} } \\ & = \lim\limits_{x\rightarrow 0} -x & & \\ & = 0. \end{align}

   Since $\ln f(x)=0, f(x) = e^{\ln f(x)} = e^0 = 1$. Hence $\lim\limits_{x\to 0}x^x=1$.



7. $\lim\limits_{x\to \frac{\pi}{2}} (\tan x)^{\cos x}$$=1$see numerical answer
   view complete solution

   Solution.

   Let $g(x)=\lim\limits_{x\to \frac{\pi}{2}} (\tan x)^{\cos x}$ $(1^\infty \text{ form})$.

   \begin{align} \ln g(x) &= \lim\limits_{x\rightarrow \frac{\pi}{2}} \cos x \ln (\tan x) & & \text{transform to } \infty \times 0~{\rm {form}} \\ & = \lim\limits_{x\rightarrow \frac{\pi}{2}} \frac{ \ln (\tan x)}{ \sec x} & & \text{transform to } \dfrac{0}{0}~{\rm {form}} \\ & = \lim\limits_{x\rightarrow \frac{\pi}{2}} \dfrac{ \dfrac{d( \ln (\tan x) )}{dx}}{ \dfrac{ d \left( \sec x \right)}{dx}} & & {\text {L'Hopital's Rule}} \\ & = \lim\limits_{x\rightarrow \frac{\pi}{2}} \dfrac{ \dfrac{\sec^2 x}{\tan x} }{ \sec x \tan x } \\ & = \lim\limits_{x\rightarrow \frac{\pi}{2}} \dfrac{\sec x}{\tan^2 x} \\ & = \lim\limits_{x\rightarrow \frac{\pi}{2}} \dfrac{\cos x}{\sin^2 x} \\ & = 0. \end{align}

   Since $\ln g(x)=0, g(x) = e^{\ln g(x)} = e^0 = 1$. Hence $\lim\limits_{x\to \frac{\pi}{2}}(\tan x)^{\cos x}=1$.

Theorem. (L'Hopital's Rule, $\frac{0}{0}$ form)

Theorem. (L'Hopital's Rule, $\frac{\infty}{\infty}$ form)

Exercises.

1. In the above proof, why does the equality relation $\textbf{(*)}$ hold?
   Show answer

   Let $\displaystyle \beta=\lim\limits_{x\to a}\frac{f(x)+g(x)}{g(x)}$. Since $\alpha=0, \beta\neq 0$.

   Therefore, we may apply the result of Case 1 on $\beta$ which suggests that $\displaystyle \lim\limits_{x\to a} \frac{f(x)+g(x)}{g(x)} = \lim\limits_{x\to a} \frac{f'(x)+g'(x)}{g'(x)}$.

2. In the above proof, why does the equality relation $\textbf{(#)}$ hold?
   Show answer

   Let $\displaystyle \gamma=\lim\limits_{x\to a}\frac{g(x)}{f(x)}$. Since $\alpha$ is infinite, $\gamma = 0$.

   Therefore, we may apply the result of Case 2 on $\gamma$ which suggests that $\displaystyle \lim\limits_{x\to a} \frac{g(x)}{f(x)} = \lim\limits_{x\to a} \frac{g'(x)}{f'(x)}$.