Question

Compute the limit
`\lim_(x \to \ 0 ) (e^(2x)-1)/(sin(x))`

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) (e^(2x)-1)/(\sin(x)) = \lim_(x\to0)(2e^(2x))/(\cos(x)) = (2e^0)/(\cos(0)) = \boxed{2}`