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'Hõpital 's rule
Let $\underset{x\to a}{\lim }(f(x){)}^{g(x)}$ is in the form of ${\infty }^0$, it can be written as $e^{\underset{x\to a}{\lim }g(x)\ell nf(x)}=e^L$
where $L=\underset{x\to a}{\lim }\frac{\ell nf(x)}{1/g(x)}$ is $\frac{\infty }{\infty }$ form and can be solved using L'Hõpital 's rule.
$\underset{x\to 1^{+}}{\mathrm{Lim}}{x}^{1/(1-x)}$

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

Answer 1

Answer: (B)

Correct answer is (b) $e^{-1}$