Question

$\displaystyle\lim_{x \to 0} \frac{xe^x - \sin \, x}{x}$ is equal to

• A. 3
• B. 1
• C. 0
• D. 2

Answer 1

Answer: (C)

0

Answer 2

$\displaystyle\lim _{x \rightarrow 0} \frac{x e^{x}-\sin\, x}{x} \,\,\left(\frac{0}{0}\text{form} \right)$
$=\displaystyle\lim _{x \rightarrow 0} \frac{x e^{x}+e^{x}-\cos\, x}{1} \,\,$ [using L' Hospital's rule]
$=0+1-1=0$