In my notes its given
$$\lim_{x\to \infty} \frac{\ln{x}}{x} = \lim_{x\to \infty} \frac{\frac{1}{x}}{1}$$
Is that correct? How do I get that?
I think another example is also related
$$x^{\frac{1}{x}} = e^{\frac{\ln{x}}{x}}$$
In my notes its given
$$\lim_{x\to \infty} \frac{\ln{x}}{x} = \lim_{x\to \infty} \frac{\frac{1}{x}}{1}$$
Is that correct? How do I get that?
I think another example is also related
$$x^{\frac{1}{x}} = e^{\frac{\ln{x}}{x}}$$
Both of your example are correct.
Please note that $\ln x\to\infty$ as $x\to\infty$. Therefore, $$\lim_{x\to\infty}=\frac{\ln x}{x}\equiv\frac{\infty}{\infty},$$ an indeterminate form. Hence by L' Hospital rule, $$\lim_{x\to\infty}=\frac{\ln x}{x}=\frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x)}=\frac{\frac{1}{x}}{1}$$
Your second example: we have $x^{\frac{1}{x}} = e^{\ln x^\frac{1}{x}}=e^{\frac{\ln x}{x}}$.