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

Question

Dr. Abubacker · Accepted Answer

(a)
$\lim_{x ightarrow 0}\frac{\left( 1 - e^{x} ight)\sin x}{(x + x^{2})x}$
$= \lim_{x ightarrow 0}\frac{\left( - x - \frac{x^{2}}{2!} - \frac{x^{3}}{3!} - ... ight)}{x(1 + x)} 	imes \lim_{x ightarrow 0}\frac{\sin x}{x}$
$= - 1 	imes 1 = - 1$

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

Explanation

Reviewed by Dr. Abubacker

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