Question

Compute the limit 29%7D" id="TexFormula1" title="\lim_{x \to \ 0 } \frac{e^{2x}-1}{sin(x)}" alt="\lim_{x \to \ 0 } \frac{e^{2x}-1}{sin(x)}" align="absmiddle" class="latex-formula">

Answer 1

Both the numerator and denominator converge to e⁰ - 1 = 0 and sin(0) = 0. Applying L'Hopital's rule gives

$\displaystyle \lim_{x\to0} \frac{e^{2x}-1}{\sin(x)} = \lim_{x\to0}\frac{2e^{2x}}{\cos(x)} = \frac{2e^0}{\cos(0)} = \boxed{2}$