# Calculus: Limits

## L'Hopital's Rule

L'Hopital's Rule is the best thing to happen to limits since the Fig Newton. It's a shortcut for evaluating limits. The only requirement is that you first know how to take derivatives.

L'Hopital's Rule: For two functions $f(x)$ and $g(x)$ defined along some interval $I$, except perhaps at a point $a$, if $\lim\limits_{x \rightarrow a} f(x) = \lim\limits_{x \rightarrow a} g(x) = 0$ or $\pm \infty$, and $g'(x) \neq 0$ where $x \neq c$, then

$$\lim\limits_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \rightarrow a} \dfrac{f'(x)}{g'(x)}$$

More plainly, if the numerator and denominator in a fraction both evaluate to either $0$ or $\infty$, then the limit of the entire fraction is equal to the limit of the derivative of the numerator over the derivative of the denominator. Note that you do not apply the quotient rule - you differentiate the numerator and denominator independently.

## Problems

1. Use L'Hopital's Rule to evaluate $\lim\limits_{x \rightarrow 0} \dfrac{\sin(x)}{x}$

$\lim\limits_{x \rightarrow 0} \dfrac{\sin(x)}{x} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}\sin(x)}{\dfrac{d}{dx}x} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\cos(x)}{1} \\ = \cos(0) \\ = 1 \\$

2. Use L'Hopital's Rule to evaluate $\lim\limits_{x \rightarrow \pi/2} \dfrac{\cos(x)}{x-\pi/2}$

$\lim\limits_{x \rightarrow \pi/2} \dfrac{\cos(x)}{x-\pi/2} \\ = \lim\limits_{x \rightarrow \pi/2} \dfrac{\dfrac{d}{dx}\cos(x)}{\dfrac{d}{dx}(x-\pi/2)} \\ = \lim\limits_{x \rightarrow \pi/2} \dfrac{-\sin(x)}{1} \\ = -\sin\left(\dfrac{\pi}{2}\right) \\ = -1 \\$

3. Determine whether L'Hopital's Rule applies to $\lim\limits_{x \rightarrow 0} \dfrac{\sin(x)}{\sin(x)}$. If so, use it to find the limit. If not, explain why.

L'Hopital's Rule applies when, upon substituting in the limiting value, the numerator and denominator both evaluate to one of 0, $+\infty$, or $-\infty$.

By substituting in $0$, we get

$\dfrac{\sin(0)}{\sin(0)} = \dfrac{0}{0}$

Thus, L'Hopital's Rule applies.

$\lim\limits_{x \rightarrow 0} \dfrac{\sin(x)}{\sin(x)} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}\sin(x)}{\dfrac{d}{dx}\sin(x)} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\cos(x)}{\cos(x)} \\ = \dfrac{\cos(0)}{\cos(0)} \\ = \dfrac{1}{1} \\ = 1 \\$

4. Determine whether L'Hopital's Rule applies to $\lim\limits_{x \rightarrow \pi/2} \dfrac{\tan(x)}{\tan\left(x+\dfrac{\pi}{2}\right)}$. If so, use it to find the limit. If not, explain why.

L'Hopital's Rule applies when, upon substituting in the limiting value, the numerator and denominator both evaluate to one of 0, $+\infty$, or $-\infty$.

By substituting in $\dfrac{\pi}{2}$, we get

$\dfrac{\tan(\pi/2)}{\tan(\pi/2 + \pi/2)} = \dfrac{\tan(\pi/2)}{\tan(\pi)} = \dfrac{\infty}{0}$

The numerator and the denominator must evaluate to the same value in order for L'Hopital's Rule to apply. Thus, we cannot use it to evaluate this limit.

5. Use L'Hopital's Rule to evaluate $\lim\limits_{x \rightarrow -2} \dfrac{x^2+4x+4}{x^2+5x+6}$. Then check it by using another method.

Part 1: Evaluate using L'Hopital's Rule:

$\lim\limits_{x \rightarrow -2} \dfrac{x^2+4x+4}{x^2+5x+6} \\ = \lim\limits_{x \rightarrow -2} \dfrac{\dfrac{d}{dx}(x^2+4x+4)}{\dfrac{d}{dx}(x^2+5x+6)} \\ = \dfrac{2x+4}{2x+5} \\ = \dfrac{2(-2)+4}{2(-2)+5} \\ = \dfrac{-4+4}{-4+5} \\ = \dfrac{0}{1} \\ = 0 \\$
Part 2: Check with another method:

$\lim\limits_{x \rightarrow -2} \dfrac{x^2+4x+4}{x^2+5x+6} \\ = \lim\limits_{x \rightarrow -2} \dfrac{(x+2)(x+2)}{(x+2)(x+3)} \\ = \lim\limits_{x \rightarrow -2} \dfrac{x+2}{x+3} \\ = \dfrac{-2+2}{-2+3} \\ = \dfrac{0}{1} \\ = 0$

6. Determine whether L'Hopital's Rule applies to $\lim\limits_{x \rightarrow 4} \dfrac{x-4}{e^{x-4}}$. If so, use it to find the limit. If not, explain why and find the limit using a different method.

L'Hopital's Rule applies when, upon substituting in the limiting value, the numerator and denominator both evaluate to one of 0, $+\infty$, or $-\infty$.

By substituting in $4$, we get

$\dfrac{4-4}{e^{4-4}} \\ = \dfrac{0}{e^0} \\ = \dfrac{0}{1} \\$

At this point we should stop and realize that we have already found the limit.

$\lim\limits_{x \rightarrow 4} \dfrac{x-4}{e^{x-4}} = 0 \\$

L'Hopital's Rule was not necessary to find this limit. But could we have used it? The answer is no. The numerator and the denominator must evaluate to the same value in order for L'Hopital's Rule to apply.

7. Use L'Hopital's Rule to evaluate $\lim\limits_{x \rightarrow 0} \dfrac{\log_2(x)}{\log_4(x)}$

$\lim\limits_{x \rightarrow 0} \dfrac{\log_2(x)}{\log_4(x)} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}\log_2(x)}{\dfrac{d}{dx}\log_4(x)} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}\frac{\ln(x)}{\ln(2)}}{\dfrac{d}{dx}\frac{\ln(x)}{\ln(4)}} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}\ln(x)}{\dfrac{d}{dx}\ln(x)}\dfrac{\ln(4)}{\ln(2)} \\ = \lim\limits_{x \rightarrow 0} 2\dfrac{\dfrac{d}{dx}\ln(x)}{\dfrac{d}{dx}\ln(x)} \\ = \lim\limits_{x \rightarrow 0} 2\dfrac{\frac{1}{x}}{\frac{1}{x}} \\ = \lim\limits_{x \rightarrow 0} 2\dfrac{x}{x} \\ = \lim\limits_{x \rightarrow 0} 2\dfrac{\dfrac{d}{dx}x}{\dfrac{d}{dx}x} \\ = \lim\limits_{x \rightarrow 0} 2\dfrac{1}{1} \\ = 2 \\$

8. Use L'Hopital's Rule to evaluate $\lim\limits_{x \rightarrow 0} \dfrac{x^3}{x^2}$

$\lim\limits_{x \rightarrow 0} \dfrac{x^3}{x^2} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}x^3}{\dfrac{d}{dx}x^2} \\ = \lim\limits_{x \rightarrow 0} \dfrac{3x^2}{2x} \\$

After applying L'Hopital's Rule, we still can't evaluate the limit. But notice that we can keep applying the rule until we can evaluate the limit.

$= \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}3x^2}{\dfrac{d}{dx}2x} \\ = \lim\limits_{x \rightarrow 0} \dfrac{6x}{2} \\ = \lim\limits_{x \rightarrow 0} 3x \\ = 3\cdot0 \\ = 0 \\$

9. Use L'Hopital's Rule to evaluate $\lim\limits_{x \rightarrow 0} \dfrac{x^a}{x^b}$ where $0<b<a$

The trick is to apply L'Hopital's Rule $b$ times such that the denominator gets turned into a constant. Mathematically, this just means taking the $b$th derivative.

$\lim\limits_{x \rightarrow 0} \dfrac{x^a}{x^b} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d^b}{dx^b}x^a}{\dfrac{d^b}{dx^b}x^b} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\frac{a!}{(a-b)!} x^{a-b}}{b!} \\ = \lim\limits_{x \rightarrow 0} \dfrac{a!}{(a-b)!b!} x^{a-b}\\ = \dfrac{a!}{(a-b)!b!} 0^{a-b} \\ = 0 \\$

10. Use L'Hopital's Rule to evaluate $\lim\limits_{x \rightarrow 0} \dfrac{\sin^2(x)}{\sin^2(x+\pi)}$

$\lim\limits_{x \rightarrow 0} \dfrac{\sin^2(x)}{\sin^2(x+\pi)} \\ = \lim\limits_{x \rightarrow 0} \dfrac{\dfrac{d}{dx}\sin^2(x)}{\dfrac{d}{dx}\sin^2(x+\pi)} \\ = \lim\limits_{x \rightarrow 0} \dfrac{2\sin(x)\cos(x)}{2\sin(x+\pi)\cos(x+\pi)} \\ = \lim\limits_{x \rightarrow 0} \dfrac{2\sin(x)\cos(x)}{2(-\sin(x))(-\cos(x))} \\ = \lim\limits_{x \rightarrow 0} \dfrac{2\sin(x)\cos(x)}{2\sin(x)\cos(x)} \\ = \lim\limits_{x \rightarrow 0} 1 \\ = 1 \\$