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