3
$\begingroup$

I saw below L'hopital Rule in a book somewhere. After I saw that, I was curious that the rule's proof is strictly correct! for example, To be the strictly proof $f\left(x\right)$, $g\left(x\right)$ must continuous differentiable?

Sorry for bad english! T^T I'm foreigner

Theorem and Proof is written in Other language Originally. So I translated them to English!

L'hopital's Theorem when $\dfrac{0}{0}$ form

Let $f\left(x\right)$, $g\left(x\right)$ be differentiable in interval $\left(a,b\right)$ that includes a point $c$ and $f\left(c\right)=g\left(c\right)=0$ and $g'\left(x\right)\neq0$ in $\left(a,b\right)$ that doesn't include a point $c$.

If ${\displaystyle \lim_{x\rightarrow c}\dfrac{f'\left(x\right)}{g'\left(x\right)}}$ exists, then ${\displaystyle \lim_{x\rightarrow c}\dfrac{f\left(x\right)}{g\left(x\right)}=\lim_{x\rightarrow c}\dfrac{f'\left(x\right)}{g'\left(x\right)}}$

p.f) At first, for $c

there exists $c$ that makes $\dfrac{f\left(x\right)-f\left(c\right)}{g\left(x\right)-g\left(c\right)}=\dfrac{f'\left(c\right)}{g'\left(c\right)}$ in $\left(c,b\right)$ at least one.

By condition, $\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{f'\left(c\right)}{g'\left(c\right)}$

thus $c

so ${\displaystyle \lim_{x\rightarrow c+}\dfrac{f\left(x\right)}{g\left(x\right)}=\lim_{d\rightarrow c+}\dfrac{f'\left(d\right)}{g'\left(d\right)}=\lim_{x\rightarrow c+}\dfrac{f'\left(x\right)}{g'\left(x\right)}}$

Alike above, thus $a

So ${\displaystyle \lim_{x\rightarrow c-}\dfrac{f\left(x\right)}{g\left(x\right)}=\lim_{d\rightarrow c-}\dfrac{f'\left(d\right)}{g'\left(d\right)}=\lim_{x\rightarrow c-}\dfrac{f'\left(x\right)}{g'\left(x\right)}}$

By above, ${\displaystyle \lim_{x\rightarrow c}\dfrac{f\left(x\right)}{g\left(x\right)}}$ equals ${\displaystyle \lim_{x\rightarrow c}\dfrac{f'\left(x\right)}{g'\left(x\right)}}$ if ${\displaystyle \lim_{x\rightarrow c}\dfrac{f'\left(x\right)}{g'\left(x\right)}}$ exists.

  • 1
    Please don't vandalize your posts. – 2017-02-05

1 Answers 1

2

You are more or less on the right track but do not use $c$ (the fixed point for the limit) as the intermediate point from MVT.

If $\lim_{x \to c}f'(x)/g'(x) = L$, then applying the mean value theorem there exists $\xi$ between $x$ and $c$ such that for $x \neq c$

$$\left|\frac{f(x)}{g(x)}- L \right|= \left|\frac{f'(\xi)}{g'(\xi)}- L \right| \to 0, $$ as $x \to c$. This follows because $\xi \to c$ as $x \to c.$

You only need $f$ and $g$ to be differentiable, not necessarily continuously differentiable.