로피탈 정리

If we plug $x=0$ into the denominator, we get 1. However, the numerator is an indeterminate form: $\tan 0 =0$, while $\displaystyle \lim_{x \to 0^+}\csc x=\infty$ and $\displaystyle \lim_{x \to 0^-}\csc x=-\infty$. If we use $\csc x = \frac{1}{\sin x}$, our expression becomes

$$\begin{aligned} \lim_{x\to0}\frac{\tan x\cdot (x^2+5)}{\sin x \cdot e^x}& \\ \text{Since plugging in $x=0$ makes both the numerator and the denominator equal to zero, this is a candidate for l'H\^ospital's Rule. However, a much easier way is to simplify the trig first.} \\ \lim_{x\to0}\frac{\tan x\cdot (x^2+5)}{\sin x \cdot e^x}&= \lim_{x\to0}\frac{\sin x\cdot (x^2+5)}{\cos x \cdot\sin x \cdot e^x}\\&= \lim_{x\to0}\frac{x^2+5}{\cos x \cdot e^x}\\ &=\frac{0^2+5}{\cos(0)\cdot e^0}=5 \end{aligned}$$