Answer
$1$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to + \infty} x^{1/x}$
Let us consider that $y=\lim\limits_{x \to + \infty} x^{1/x} \implies \ln y=\lim\limits_{x \to + \infty} \dfrac{\ln (x)}{x}$
Apply L-Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\ln y= \lim\limits_{x \to + \infty} \dfrac{1}{x} \\=0$
Thus, we have: $ y=e^0=1$
