Why is $ \lim_{x \to \infty} \ x^{2/x} = 1$ Why is $\displaystyle \lim_{x \to \infty} \ x^{2/x} = 1$ since this is an indeterminate form $\infty^{0}$ and I can't see any manipulation that would suggest this result?

If $f(x)=x^{\frac{2}{x}}$, then $\ln f(x)=\ln x^{\frac{2}{x}}=\frac{2}{x}\ln x$, and
$$\lim_{x\rightarrow \infty}\ln f(x)=\lim_{x\rightarrow \infty}\frac{2}{x}\ln x =\lim_{x\rightarrow \infty}\frac{2\ln x}{x},$$
and the indetermination now is in the form $\frac{\infty}{\infty}$. Then, for L'Hospital rule:
$$\lim_{x\rightarrow \infty}\ln f(x)=\lim_{x\rightarrow \infty}\frac{2\ln x}{x}=\lim_{x\rightarrow \infty} \frac{2\frac{1}{x}}{1}=0,$$
then
$$0=\lim_{x\rightarrow \infty}\ln f(x)=\ln \lim_{x\rightarrow \infty} f(x). $$
Now, $\ln w=0\Leftrightarrow w=1$, then
$$ \lim_{x\rightarrow \infty} f(x)=1,$$
i.e.,
$$ \lim_{x\rightarrow \infty} x^{\frac{2}{x}}=1.$$
