Question

The value of ${lim}_{x\to 0}({\left(\sin x\right)}^{1/x}+{(1+x)}^{\sin x})$ whre $x>0$ is

• A. $0$
• B. $-1$
• C. $1$
• D. 2

Answer 1

The correct option is A $0$

Given ${lim}_{x\to 0}({\left(sinx\right)}^{\frac{1}{x}}+{(1+x)}^{sinx})$

$={lim}_{x\to 0}{\left(sinx\right)}^{\frac{1}{x}}+{lim}_{x\to 0}{(1+x)}^{sinx}$

$={\left(sin0\right)}^{\frac{1}{0}}+{lim}_{x\to 0}{(1+x)}^{sinx}$

$={\left(0\right)}^{\infty }+{lim}_{x\to 0}{(1+x)}^{sinx}$

$=0+e^{{lim}_{x\to 0}\left[log{(1+x)}^{sinx}\right]}$

$(\because e^{loga}=a)$

$=e^{{lim}_{x\to 0}[sinx.log(1+x)]}$

$=e^{{lim}_{x\to 0}sinx\times {lim}_{x\to 0}log(1+x)}$

$=e^{sin\left(0\right)\times log(1+0)}$

$=e^{0\times log\left(1\right)}$

$=e^{0\times 0}=e^0$

$=1$