📂Calculus

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 $(a,x)$.

Cauchy’s Mean Value Theorem

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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