L'Hôpital's Rule/Proof 1
Theorem
Let $f$ and $g$ be real functions which are differentiable on the open interval $\openint a b$.
Let:
- $\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the derivative of $g$ with respect to $x$.
Let:
- $\ds \lim_{x \mathop \to a^+} \map f x = \lim_{x \mathop \to a^+} \map g x = 0$
Then:
- $\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
provided that the second limit exists.
Proof
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Let $l = \ds \lim_{x \mathop \to a^+} \frac {\map {f'} x}{\map {g'} x}$.
Let $\epsilon \in \R_{>0}$.
By the definition of limit, we ought to find a $\delta \in \R_{>0}$ such that:
- $\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map f x} {\map g x} - l} < \epsilon$
Fix $\delta$ such that:
- $\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map {f'} x} {\map {g'} x} - l} < \epsilon$
which is possible by the definition of limit.
Define:
- $\map {f_0} x = \begin{cases}
\map f x & : x \in \openint a b \\ 0 & : x = a \end{cases}$
- $\map {g_0} x = \begin{cases}
\map f x & : x \in \openint a b \\ 0 & : x = a \end{cases}$
By definition of right-continuous, it follows that $f_0$ and $g_0$ are continuous on $\hointr a b$.
Let $x_\delta$ be such that $0 < x_\delta - a < \delta$.
We have that $f_0$ and $g_0$ are continuous on $\closedint a {x_\delta}$, and differentiable on $\openint a {x_\delta}$
Thus, by the Cauchy Mean Value Theorem with $b = x_\delta$:
- $\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'_0} \xi} {\map {g'_0} \xi} = \dfrac {\map {f_0} {x_\delta} - \map {f_0} a} {\map {g_0} {x_\delta} - \map {g_0} a}$
Since $\map {f_0} a = \map {g_0} a = 0$:
- $\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'_0} \xi} {\map {g'_0} \xi} = \dfrac {\map {f_0} {x_\delta}} {\map {g_0} {x_\delta}}$
But since $\xi, x_\delta \in \openint a b$:
- $\map {f'_0} \xi = \map {f'} \xi$
- $\map {g'_0} \xi = \map {g'} \xi$
- $\map {f_0} {x_\delta} = \map f {x_\delta}$
- $\map {g_0} {x_\delta} = \map g {x_\delta}$
Therefore:
- $\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f {x_\delta}} {\map g {x_\delta}}$
Now, as $a < \xi < x_\delta$, it follows that $\size{\xi - a} < \delta$ as well.
Therefore:
- $\size {\dfrac {\map f {x_\delta}} {\map g {x_\delta}} - l} = \size {\dfrac {\map {f'} \xi} {\map {g'} \xi} - l} < \epsilon$
which leads us to the desired conclusion that:
- $\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
$\blacksquare$