Question

$\underset{x\to 0^{+}}{lim}{x}^x$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $\infty$

Answer 1

The correct option is B $1$

Let, $L=\underset{x\to 0^{+}}{lim}{x}^x$
Taking $\log$ both side,
$\log L=\underset{x\to 0^{+}}{lim}x\log x=\underset{x\to 0^{+}}{lim}\frac{\log x}{\frac{1}{x}},\frac{\infty }{\infty }$ form
Thus using L-hospital rule,
$\log L=\underset{x\to 0^{+}}{lim}\frac{\frac{1}{x}}{\frac{-1}{x^2}}=\underset{x\to 0^{+}}{lim}(-x)=0$
$\Rightarrow L=1$