Proof of L'Hôpital's Rule
Theorem1
Given that $f(x)$ and $g(x)$ are differentiable near $x=a$ and that $g ' (x) \ne 0$ and $\displaystyle \lim _{x \to a} f(x) = \lim _{x \to a} g(x) = 0$,
$$ \lim _{x \to a} {{f(x)} \over {g(x)}} = \lim _{x \to a} {{f ' (x)} \over {g ' (x)}} $$
Explanation
Though this theorem may seem like a magic wand to many students, who have learned and applied it extensively, personally, I believe it’s better to stick to the basics and seal this knowledge away until a few months before the college entrance exam.
Interestingly, the first proof of this theorem was not given by L’Hôpital but by the mathematician Johann Bernoulli, whom L’Hôpital had patronized.
Proof
Since $f(a)=g(a)=0$,
$$ \lim _{x \to a} {{f(x)} \over {g(x)}} = \lim _{x \to a} {{f(x)-f(a)} \over {g(x)-g(a)}} $$
Also,
$$ f(x)-f(a) = \begin{cases} f(x) & ,x \ne a \\ 0 & , x=a \end{cases} \\ g(x)-g(a) = \begin{cases} g(x) & ,x \ne a \\ 0 & , x=a \end{cases} $$
Therefore, ${f(x)-f(a)}$ and ${g(x)-g(a)}$ are continuous at $[x,a]$ or $[a,x]$ and differentiable at $(x,a)$ or $(a,x)$.
If function $f(x), g(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and if $g ' (x) \ne 0$, then there exists at least one $c$ in $(a,b)$ that satisfies $\displaystyle {{f ' (c)}\over{g ' (c)}}={{f(b)-f(a)}\over{g(b)-g(a)}}$.
By Cauchy’s Mean Value Theorem, there exists at least one $c$ in $(x,a)$ or $(a,x)$ that satisfies $\displaystyle {{f ' (c)}\over{g ' (c)}}={{f(x)-f(a)}\over{g(x)-g(a)}}$. Since $c$ exists in $(x,a)$ or $(a,x)$, when $x \to a$, then $c \to a$, and
$$ \begin{align*} \lim _{x \to a} {{f(x)} \over {g(x)}} =& \lim _{x \to a} {{f(x)-f(a)}\over{g(x)-g(a)}} \\ =& \lim _{c \to a} {{f ' (c)}\over{g ' (c)}} \\ =& \lim _{x \to a} {{f ' (x)}\over{g ' (x)}} \end{align*} $$
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James Stewart, Daniel Clegg, and Saleem Watson, Calculus (early transcendentals, 9E), pA48-A49 ↩︎