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 $li{m}_{x\to 1^{+}}{x}^{1/(1-x)}$

• A. -1
• B. $e^{-1}$
• C. -2
• D. $e^{-2}$

Answer 1

Answer: (B)

$e^{-1}$

Let,$L=\underset{x\to 1^{+}}{lim}{x}^{\frac{1}{1-x}}>0$
Taking log on both sides,
$\log L=\underset{x\to 1^{+}}{lim}\frac{\log x}{1-x}$
Clearly ,the limit is of the form $\frac{0}{0}$,
Applying L'Hopital's rule,
$\log L=\underset{x\to 1^{+}}{lim}\frac{1}{-x}=-1$
$\Rightarrow L=e^{-1}$