Answer
$1$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\ln f(x)=x \ln x$
Now, $\lim\limits_{x \to 0} f(0)=\dfrac{\infty}{\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{x \to 0} \frac{1/x}{-1/x^2}}=e^{\lim\limits_{x \to 0} (-x)}$
Thus,
$e^{0}=1$
