Question

If $I=\underset{x\to 0}{\lim }\sin \left(\frac{e^x-x-1-\frac{x^2}{2}}{x^2}\right)$, then limit

• A. does not exist
• B. exists and equals 1
• C. exists and equals 0
• D. exists and equals $\frac{1}{2}$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\sin \left(\frac{e^x-x-1-\frac{x^2}{2}}{x^2}\right)$
$=\underset{x\to 0}{\lim }\frac{e^x-x-1-\frac{x^2}{2}}{x^2}$
$=\underset{x\to 0}{\lim }\frac{e^x-1-0-x}{2x}$
$=\underset{x\to 0}{\lim }\frac{e^x-1}{2}=0$