Answer
1
Work Step by Step
$\lim\limits_{x \to 0}$$\frac{e^x-1}{tanx}$
After plugging in 0 for x in the numerator and denominator, we get $\frac{0}{0}$, an indeterminate form. This is one of the forms where we can use L'Hopital's Rule.
We take the derivative of the top and the bottom and get:
$\lim\limits_{x \to 0}$$\frac{e^x}{(sec^2x)}$
We then plug in 0 again, and we get:
$\frac{e^0}{(sec0)^2}$ = $\frac{1}{1}$ = 1
