Extension: Proving the 0/0 Case of L'Hopital

L'Hospital's rule provides a method for finding limits for expressions that would otherwise be impossible to calculate. The most basic form of the rule says that if $f(x)$ and $g(x)$ are both differentiable at $x=a$, and $$\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = 0,$$ then $\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}$.

Let's prove this rule. Recall our method of locally linear approximations for differentiable functions: $f(a) + f'(a)(x-a) \to f(x)$ as $x \to a$. Thus, we can rewrite the desired limit as $$\lim_{x \to a} \left(\frac{f(a) + f'(a)(x-a)}{g(a) + g'(a)(x-a)}\right).$$ Furthermore, as the functions are both continuous, $f(a) = g(a) = 0$, and thus the limit becomes

$$\lim_{x \to a} \frac{f(a) + f'(a)(x-a)}{g(a) + g'(a)(x-a)} = \lim_{x \to a} \frac{f'(a)(x-a)}{g'(a)(x-a)} = \frac{f'(a)}{g'(a)}.$$

This is L'Hospital's rule, and it allows us to evaluate limits of indeterminate forms! Let's try a familiar example: what is $\lim_{x \to 0} \frac{x^2 + x}{x}$? Taking the derivatives of the numerator and denominator: we get $\frac{2x+1}{1} = 1$. But there are other kinds of indeterminate forms. How would we find the limit of a function that converges to $1^\infty$? More concretely, let $L = \lim_{x \to 0} (e^x + x)^{\frac{1}{x}}$.

If we take the natural logarithm of both sides, we get

$$\begin {align*} \ln L &= \lim_{x \to 0} \frac{\ln(e^x + x)}{x} \\&= \lim_{x \to 0} \frac{\left(\frac{e^x + 1}{e^x + x}\right)}{1} \\&= 2, \end {align*}$$ and $L = e^2$.

There are many other types of indeterminate forms, such as $0^0, \infty^0$, and $\infty - \infty$, but if you can manipulate them into the form $\frac{0}{0}$, you can use this method to evaluate them.

Let $f(x)$ and $g(x)$ be continuous differentiable functions that satisfy $f(a)=g(a)=0$. Then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$

Proof:

$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}=\lim_{x\to a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$

This works because $f(a)=g(a)=0$, so subtracting these values does not change the numerator or denominator in the first step.