Answer
$a)$ $\infty^{0}$
$b)$ $1$
$c)$ see picture
Work Step by Step
$a)$ $\lim\limits_{x \to \infty}x^{\frac{1}{x}}=\infty^{\frac{1}{\infty}}=\infty^{0}$
$\infty^{0}$ is an indeterminate form.
$b)$ $\lim\limits_{x \to \infty}x^{\frac{1}{x}}=\infty^{\frac{1}{\infty}}=\infty^{0}$
$\infty^{0}$ is an indeterminate form. We can't do l'Hopital's Rule until the indeterminate form is either $\frac{0}{0}$ or $\frac{\infty}{\infty}$, so:
$let$ $y=\lim\limits_{x \to \infty}x^{\frac{1}{x}}$
$lny=\lim\limits_{x \to \infty}(\frac{1}{x})lnx$
$lny=\lim\limits_{x \to \infty}\frac{lnx}{x}$
$lny=\lim\limits_{x \to \infty}\frac{ln\infty}{\infty}$
$lny=\lim\limits_{x \to \infty}\frac{\infty}{\infty}$
Now we can do l'Hopital's Rule:
$lny=\lim\limits_{x \to \infty}(\frac{\frac{1}{x}}{1})$
$lny=\lim\limits_{x \to \infty}\frac{1}{x}$
$lny=\lim\limits_{x \to \infty}\frac{1}{\infty}$
$lny=0$
$e^{lny}=e^{0}$
$y=1$
$c)$ see picture