How can I prove that the following are happening: $lnBig(1+frac{1}{x}Big)=frac{1}{x}+oBig(frac{1}{x}Big)$?

Among various choice of $x_0$, we assume $x_0=infty$ because otherwise your question is false.

Explicitly, we want to show $lim_{xrightarrowinfty}x(ln(1+frac{1}{x})-frac{1}{x})=0$.

Putting $y=frac{1}{x}$, it suffices to show $lim_{yrightarrow 0}frac{ln(1+y)}{y}=1$.

This can be written as $lim_{yrightarrow 0}frac{ln(1+y)-ln 1}{y-0}=1Leftrightarrow ln'(1)=1$ and we are done.

Using Taylor's formula, you know that $$ log(1+y) = y + o(y), quad (y to 0) $$

Setting $y=1/x$ gives you

$$ logleft(1+frac 1xright) = frac 1x + oleft(frac 1xright), quad (xto +infty) $$

The same can be accomplished in the second example:

$$ (1+y)^p = 1 + p y + o(y) (textrm{ as }y to 0) Rightarrow left(1+frac 1xright)^p = 1+ frac px + oleft(frac 1x right) (textrm{ as } xto +infty) $$