L'Hôpital's rule

L'Hôpital's rule is a method that lets us use derivatives in evaluating limits involving "indeterminate forms," i.e. when a straight-forward approach gives us $\dfrac{0}{0}$ or $\dfrac{\pm \infty}{\pm \infty}$. $$ $$ More specificially, l'Hôpital's rule tells us that when $$\dfrac{\lim\limits_{x \to a}\ f(x) }{\lim\limits_{x \to a}\ g(x) } = \dfrac{0}{0} \text{ or } \dfrac{\pm \infty}{\pm \infty},$$ $$\lim\limits_{x \to a}\ \dfrac{f(x) }{ g(x) } = \lim\limits_{x \to a}\ \dfrac{f'(x) }{ g'(x) },$$ where the primes ( ' ) signify taking the derivative with respect to $x$.

Since $\dfrac{\sin 0}{0} = \dfrac{0}{0},$ we apply l'Hôpital's rule: \begin{eqnarray} \lim\limits_{x \to 0}\ \dfrac{\sin x}{x} &=& \lim\limits_{x \to 0}\ \dfrac{(\sin x)'}{(x)'} \\ \ &=& \lim\limits_{x \to 0}\ \dfrac{\cos x}{1} \\ \ &=& \cos 0 \\ \ &=& 1. \end{eqnarray}

Since $$\dfrac{\lim\limits_{x \to -3}\ (x + 3)^3}{\lim\limits_{x \to -3}\ (x^2 + 9)} = \ \dfrac{0}{0},$$ we apply l'Hôpital's rule: \begin{eqnarray} \lim\limits_{x \to -3}\ \dfrac{(x + 3)^3}{x^2 + 9} &=& \lim\limits_{x \to -3}\ \dfrac{3(x + 3)^2}{2x} \\ \ &=& \dfrac{0}{-6} \\ \ &=& 0. \end{eqnarray}

Since $$\dfrac{\lim\limits_{x \to \infty}\ e^x}{\lim\limits_{x \to \infty}\ (x^2 + 4)} = \dfrac{\infty}{\infty},$$ we apply l'Hôpital's rule, which gives us \begin{eqnarray} \lim\limits_{x \to \infty}\ \dfrac{e^x}{x^2 + 4} &=& \lim\limits_{x \to \infty}\ \dfrac{(e^x)'}{(x^2 + 4)'} \\ \ &=& \lim\limits_{x \to \infty}\ \dfrac{e^x}{2x}. \end{eqnarray} Since $$\dfrac{\lim\limits_{x \to \infty}\ e^x}{\lim\limits_{x \to \infty}\ 2x} = \dfrac{\infty}{\infty},$$ we apply l'Hôpital's rule a second time: \begin{eqnarray} \lim\limits_{x \to \infty}\ \dfrac{e^x}{2x} &=& \lim\limits_{x \to \infty}\ \dfrac{(e^x)'}{(2x)'} \\ \ &=& \lim\limits_{x \to \infty}\ \dfrac{e^x}{2} \\ \ &=& \dfrac{\infty}{2} \\ \ &=& \infty. \end{eqnarray} And so, the answer is $$\lim\limits_{x \to \infty}\ \dfrac{e^x}{x^2 + 4} = \infty.$$