Limit prob

$\begin{aligned} L & = \lim_{x \to 0} \frac {(e^x-1)\tan^2 x}{x^3} \\ & = \lim_{x \to 0} \frac {e^x-1}x \left(\frac {\tan x}x\right)^2 & \small {\color{#3D99F6}\text{Both are 0/0 cases, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0} \frac {e^x}1 \left(\frac {\sec^2 x}1 \right)^2 & \small {\color{#3D99F6}\text{Differentiate up and down}} \\ & = \boxed{1} \end{aligned}$

Without L'Hopital's rule:

Note that $\displaystyle \lim_{x \to 0} \dfrac{\sinh x}{x} = 1$ and $\displaystyle \lim_{x\ to 0} \dfrac{\sin x}{x} =1$ .

$\sinh x = \dfrac{e^x - e^{-x} }{2} = \dfrac{e^{-x} (e^{2x} -1 )}{2} = \dfrac{e^{-x}(e^x + 1)(e^x - 1)}{2}$

Thus, what I'm going to do is rewrite the limit as:

$\displaystyle \lim_{x \to 0} \dfrac{e^{-x}(e^x + 1)(e^x - 1)}{2} \cdot \dfrac{\sin^2 x}{x^3 \cos^2 x} \cdot \dfrac{2}{e^{-x} (e^x + 1)}$

$=\displaystyle \lim_{x \to 0} \dfrac{\sin^2 x \sinh x}{x^3 \cos^2 x} \cdot \dfrac{2}{e^{-x} (e^x + 1)}$

$=\displaystyle \lim_{x \to 0} \dfrac{\sin x }{x} \cdot \dfrac{\sin x }{x} \cdot \dfrac{\sinh x}{x} \cdot \dfrac{1}{\cos^2 x} \cdot \dfrac{2}{e^{-x} (e^x + 1)}$

Break it up into a product of five limits, and the first four all exist and equal one, and the last, upon substituting in $x = 0$ , we get $\boxed{1}$