$\lim _{x \rightarrow 0} \frac{\left(1-e^x\right) \sin x}{x^2+x^3}$ is equal to

Question

Aicharya · Accepted Answer

$\begin{aligned} & \lim _{x ightarrow 0} \frac{\left(1-e^xight) \sin x}{\left(x+x^2ight) x} \ & =\lim _{x ightarrow 0} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldotsight)}{x(1+x)} 	imes \lim _{x ightarrow 0} \frac{\sin x}{x} \ & =-1 	imes 1=-1\end{aligned}$

$\lim _{x \rightarrow 0} \frac{\left(1-e^x\right) \sin x}{x^2+x^3}$ is equal to

Solution