LHopitalsrule

L'Hopital's Rule

L'Hôpital's Rule is a powerful mathematical tool used for finding limits of indeterminate forms of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$. According to this rule:

If

$\lim_{x \to a} f(x) = 0$

and

$\lim_{x \to a} g(x) = 0$

or both limits are infinite, and the functions $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$), then

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

provided the limit on the right exists or is infinite.

Proof of L'Hopital's Rule:

Proof for the form $\frac{0}{0}$:

Given that $f(a) = 0$ and $g(a) = 0$, let's assume $g'(a) \neq 0$ and that $f(x)$ and $g(x)$ are differentiable on an interval $I$ containing $a$ except possibly at $a$.

Now, by the definition of the derivative,

$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

and

$g'(a) = \lim_{x \to a} \frac{g(x) - g(a)}{x - a}$

Since $f(a) = g(a) = 0$, these become:

$f'(a) = \lim_{x \to a} \frac{f(x)}{x - a}$

and

$g'(a) = \lim_{x \to a} \frac{g(x)}{x - a}$

Now, using Cauchy's Mean Value Theorem, there exists some $c$ in $(a, x)$ such that:

$\frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c)}{g'(c)}$

But $f(a) = 0$ and $g(a) = 0$. So:

$\frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}$

As $x \to a$, $c$ also approaches $a$. So, taking the limit as $x \to a$ on both sides, we get:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{c \to a} \frac{f'(c)}{g'(c)} = \frac{f'(a)}{g'(a)}$

This concludes the proof for L'Hopital's Rule for the case $\frac{0}{0}$. The proof for the $\frac{\infty}{\infty}$ form is analogous.

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This explanation and proof are simplified and rely on the Mean Value Theorem. L'Hôpital's Rule is a profound result that helps simplify the evaluation of many difficult limits.