Question

$\underset{x\to 0}{lim}\frac{(1-e^x)\sin x}{x^2+x^3}=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$-1$

The expression can be simplified as,
(lim $x\to 0\frac{(1-e^x)}{x})\times (\frac{sinx}{x}\left)\right(li{m}_{x\to 0}\frac{1}{1+x})$
Now using,
lim $x\to 0\frac{(e^x-1)}{x}=1$
lim $x\to 0\frac{sinx}{x}$ $=1$
We get,
$(-1)$ $\times 1\times limx\to 0\frac{1}{1+x}$
$=-1$