Question

Solve:

$li{m}_{x\to 0}\frac{e^{sinx}-1}{x}.$

Answer 1

${lim}_{x⟶0}\frac{e^{\sin x}-1}{x}={lim}_{x⟶0}\frac{e^{\sin x}-1}{\sin x}.\frac{\sin x}{x}={lim}_{x⟶0}\frac{e^{\sin x}-1}{\sin x}{lim}_{x⟶0}\frac{\sin x}{x}={lim}_{\sin x⟶0}\frac{e^{\sin x}-1}{\sin x}{lim}_{x⟶0}\frac{\sin x}{x}(\because whenx⟶0,\sin x⟶0)=1.1=1(\because {lim}_{x⟶0}\frac{\sin x}{x}=1and{lim}_{x⟶0}\frac{e^{\sin x}-1}{\sin x}=1)$