I recently tried to prove L'Hospitals Rule on my own but I'm not sure if it is correct because it differs from all other proofs that I have already found.
Here's the proof:
Let functions $f,g:(a,b)\rightarrow \mathbb{R}$ be continuous on $(a,b)$ except possibly at $x_{0}$ $\epsilon$ $(a,b)$.If:
- $\lim_{x \rightarrow x_{0}} f(x)=0$ and $\lim_{x \rightarrow x_{0}} g(x)=0$
- ($\forall x $ $\epsilon$ $ (a,b)\setminus$ {$x_{0}$}) $\exists f′(x)$ and $\exists g′(x)$ which is diferent from zero
- $\exists lim_{x \rightarrow x_{0}}\frac{f′(x)}{g′(x)}=A$ where $A$ $\epsilon$ $\mathbb{R}$ $\cup$ {$-\infty,+\infty$}
Then there exists $lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}$ and $lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}=A$
Proof:
- Let $f(x_{0})=g(x_{0})=0$ , and $f′, g′$ be continuous on $(a,b)$ , then :
$lim_{x \rightarrow x_{0}}\frac{f′(x)}{g′(x)}=\frac{f′(x_{0})}{g′(x_{0})}=\frac{lim_{x \rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}}{lim_{x \rightarrow x_{0}}\frac{g(x)-g(x_{0})}{x-x_{0}}}=lim_{x \rightarrow x_{0}}\frac{f(x)-f(x_{0})}{g(x)-g(x_{0})}=lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}$
- General proof (regardless of the values $f(x_{0})$,$g(x_{0})$)
Let's consider the functions $F(x)$ and $G(x)$ such that :
$F(x)=\begin{cases} f(x) &\text{if x is different from } x_{0} \\{}\\ 0 &\text{ for } x=x_{0} .\end{cases}$
$G(x)=\begin{cases} g(x) &\text{if x is different from } x_{0} \\{}\\ 0 &\text{ for } x=x_{0} .\end{cases}$
It is obvious that $F(x)$ and $G(x)$ are continuous at $x_{0}$ $\epsilon$ $(a,b)$.
Let $x>x_{0}$ , this means that $F(x)$ and $G(x)$ are continuous on $[x_{0},x]$, differentiable at least on $(x_{0},x)$ and $\forall$ $x_{1}$ $\epsilon$ $(x_{0},x)$ $\exists$ $F′(x)$ and $\exists G′(x)$ which is different from zero. So this means that functions $F(x),G(x)$ satisfy the conditions of Cauchy's Mean Value Theorem therefore there $\exists$ $c$ $\epsilon$ $(x_{0},x)$ such that:
$\frac{f′(c)}{g′(c)}=\frac{F′(c)}{G′(c)}=\frac{F(x)-F(x_{0})}{G(x)-G(x_{0})}=\frac{F(x)}{G(x)}=\frac{f(x)}{g(x)}$
(Because $c$ and $x$ are different from $x_{0}$)
Now from $c$ $\epsilon$ $ (x_{0},x)$ when $x \rightarrow x_{0}^+$ $\implies$ $c\to x_{0}^+$
Thus we have:
$lim_{x \rightarrow x_{0}^+}\frac{f′(x)}{g′(x)}=lim_{c \rightarrow x_{0}^+} \frac{f′(c)}{g′(c)}=lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}$
Analogously we prove when $x\to x_{0}^-$
When $x\to \infty$ we define function $f(x)$ and $g(x)$ be continuous on $(a,\infty)$ and substitute $x=\frac{1}{t}$, so we'll have:
$lim_{x \rightarrow \infty}\frac{f(x)}{g(x)}=lim_{t \to 0^+} \frac{f(\frac{1}{t})}{g(\frac{1}{t})}=lim_{t \rightarrow 0^+} \frac{\big(f(\frac{1}{t})\big)′}{\big(g(\frac{1}{t})\big)′}=lim_{t \rightarrow 0^+}\frac{\frac{-1}{t^2} f′(\frac{1}{t})}{\frac{-1}{t^2}g′(\frac{1}{t})}=lim_{t\rightarrow 0^+} \frac{f′(\frac{1}{t})}{g′(\frac{1}{t})}=lim_{x \to \infty} \frac{f′(x)}{g′(x)}=A$
When $x \to -\infty$ then we take $t \to 0^-$
If someone would tell me if my proof is correct or not I'd really appreciate it.Thank you very much!
Your proof is not correct. Early in the proof you assert that if you let $f(x_0)=g(x_0)=0$ then $f$ and $g$ (you claim $f'$ and $g'$ which is also false, but I think you meant $f$ and $g$) are continuous and differentiable on $(a,b)$. But this isn't the case. They weren't necessarily continuous at $x_0$ to start with. Substituting $0$ for their value there won't make them continuous much less differentiable.