## Calculus: Early Transcendentals 8th Edition

$\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