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Q. $\displaystyle\lim _{x \rightarrow 0} x^{x}$ is equal to

Limits and Derivatives

Solution:

Let $y=\displaystyle\lim _{x \rightarrow 0} x^{x}$
$\Rightarrow \log y=\displaystyle\lim _{x \rightarrow 0} x \log x$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\log x}{1 / x} \,\,\,\ \left(\frac{\infty}{\infty}\right.$ form $)$
$=\displaystyle\lim _{x \rightarrow 0} \frac{1 / x}{-1 / x^{2}}$
$=-\displaystyle\lim _{x \rightarrow 0} x=0$
$\Rightarrow y=e^{0}=1 .$