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proof of De l’Hôpital’s rule (TeX) (PDF)
$\newcommand{\R}{\mathbb R}$
Let $x_0\in \R$, $I$ be an interval containing $x_0$ and let $f$ and $g$ be two differentiable functions defined on $I\setminus\{x_0\}$ with $g'(x)\neq 0$ for all $x\in I$. Suppose that
\[
\lim_{x\to x_0} f(x) = 0, \quad \lim_{x\to x_0} g(x)=0
\]
and that
\[
\lim_{x\to x_0} \frac{f'(x)}{g'(x)}=m.
\]
We want to prove that hence $g(x)\neq 0$ for all $x\in I\setminus\{x_0\}$ and
\[
\lim_{x\to x_0} \frac{f(x)}{g(x)}=m.
\]
First of all (with little abuse of notation) we suppose that $f$ and $g$ are defined also in the point $x_0$ by $f(x_0)=0$ and $g(x_0)=0$. The resulting functions are continuous in $x_0$ and hence in the whole interval $I$.
Let us first prove that $g(x)\neq 0$ for all $x\in I\setminus\{x_0\}$. If by contradiction $g(\bar x)=0$ since we also have $g(x_0)=0$, by Rolle's Theorem we get that $g'(\xi)=0$ for some $\xi\in (x_0,\bar x)$ which is against our hypotheses.
Consider now any sequence $x_n\to x_0$ with $x_n\in I\setminus\{x_0\}$.
By Cauchy's mean value Theorem there exists a sequence $x'_n$ such that
\[
\frac{f(x_n)}{g(x_n)} = \frac{f(x_n)-f(x_0)}{g(x_n)-g(x_0)}
= \frac{f'(x'_n)}{g'(x'_n)}.
\]
But as $x_n\to x_0$ and since $x'_n \in (x_0,x_n)$ we get that $x'_n\to x_0$ and hence
\[
\lim_{n\to\infty} \frac{f(x_n)}{g(x_n)}
= \lim_{n\to\infty} \frac{f'(x_n)}{g'(x_n)}
= \lim_{x\to x_0} \frac{f'(x)}{g'(x)} = m.
\]
Since this is true for any given sequence $x_n\to x_0$ we conclude that
\[
\lim_{x\to x_0} \frac{f(x)}{g(x)} = m.
\]
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hbghlyj
发表于 2024-2-20 21:09
