Question

Limits that lead to the indeterminate forms $1^{\infty },0^0,{\infty }^0$ can sometimes be solved taking logarithm first and then using L' Hospital's rule

Let $\underset{x\to a}{lim}\left(f\right(x){)}^{g\left(x\right)}$ is in the form of ${\infty }^0$, it can be written as $e^{{lim}_{x\to a}g\left(x\right)\ln f\left(x\right)}=e^L$ where $L=\underset{x\to a}{lim}\frac{\ln f\left(x\right)}{\frac{1}{g\left(x\right)}}$ is $\frac{\infty }{\infty }$ form and can be solved using L' Hopital's rule.

Then $\underset{x\to 0^{+}}{lim}(sinx{)}^{2sinx}$ :

• A. $1$
• B. $0$
• C. $2$
• D. does not exist

Answer 1

The correct option is A $1$

Let,$m=\underset{x\to 0^{+}}{lim}(sinx{)}^{2sinx}$

Taking $\log$ on both sides, we get

$\Rightarrow \log m=\underset{x\to 0^{+}}{lim}2sinx\log \left(sinx\right)$

$\Rightarrow \log m=\underset{x\to 0^{+}}{lim}\frac{2\log \left(sinx\right)}{cosecx}$

Clearly, the limit is of the form$\frac{\infty }{\infty }$, so applying L- Hospital's 's rule, we get

$\Rightarrow \log m=\underset{x\to 0^{+}}{lim}-\frac{2cosx}{cosecxcotxsinx}$

$\Rightarrow \log m=\underset{x\to 0^{+}}{lim}-\frac{2cosx}{cotx}$

$\Rightarrow \log m=0$

$\Rightarrow m=1$

$=\underset{x\to 0^{+}}{lim}(sinx{)}^{2sinx}=1$