Answer
$1$
Work Step by Step
Here, we have $\ln f(x)=x \ln (1+\dfrac{1}{x})$
Now, $e^{\lim\limits_{x \to 0} x \ln (1+\frac{1}{x})}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$e^{[\lim\limits_{x \to 0} \dfrac{-1/x(x+1)}{-1/x^2}]}=e^{[\lim\limits_{x \to 0} \dfrac{x}{x+1}]}$
and $e^{\frac{0}{0+1}}=1$
