> [!theorem] > > Let $f$, $g$ be [[Continuity|continuous]] [[Function|functions]] defined on an interval containing $a$. If $f$ and $g$ are [[Derivative|differentiable]] on this interval, and that $f(a) = g(a) = 0$, then > $ > \lim_{x \to a}\frac{f^\prime(x)}{g^\prime(x)} = L \Rightarrow > \lim_{x \to a}\frac{f(x)}{g(x)} = L > $ > > *Proof*. > $ > \begin{align*} > \lim_{x \to a}\frac{f^\prime(x)}{g^\prime(x)} &= > \frac{f^\prime(x)}{g^\prime(x)} \\ > &= > \frac{ > \lim_{x \to a}\frac{f(x) - f(a)}{x - a} > }{ > \lim_{x \to a}\frac{g(x) - g(a)}{x - a} > } \\ > &= > \lim_{x \to a}\frac{f(x) - f(a)}{g(x) - g(a)} = > \lim_{x \to a}\frac{f(x)}{g(x)} > \end{align*} > $ > [!summary] Other Indeterminate Forms > > Many other indeterminate forms can be transformed into a $\frac{0}{0}$ form or an $\frac{\infty}{\infty}$ form, and then evaluated using L'Hôpital's Rule. > > > #### Indeterminate Products > > For any product $f \cdot g$ that results in the indeterminate form $0 \cdot \pm\infty$, the limit can be transformed to $\frac{f}{1/g}$, converting it to a $\frac{0}{0}$ form. > > > #### Indeterminate Differences > > For any difference $f - g$ that results in some form of $\infty - \infty$, it can be converted into an indeterminate quotient by multiplying it by a fraction of its conjugate: > $ > \lim(f - g) = \lim\frac{f^2 - g^2}{f + g} > $ > > > #### Indeterminate Powers > > There can be three types of indeterminate powers that can arise from $f^g$: $0^0$, $\infty^0$, and $1^\infty$. All three can be rewritten to be an [[Exponential Function|exponential function]]: $e^{g\ln{f}}$. The resulting indeterminate product in the exponent can then be converted into a quotient.