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

Question

How can I prove that the following are happening ($x\to\infty$): $\ln\Big(1+\frac{1}{x}\Big)=\frac{1}{x}+o\Big(\frac{1}{x}\Big)$ and $\Big(1+\frac{1}{x}\Big)^{p}=1+\frac{p}{x}+o\Big(\frac{1}{x}\Big)$, where o is the notation for Little-o. I thought it could be shown directly with the definition that 2 functions are asymptotically equivalent ($\lim_{x\to x_0}\vert\frac{f(x)}{g(x)}\vert$=0), but I'm not sure.

Answer 1

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

Setting $y=1/x$ gives you

$$ \log\left(1+\frac 1x\right) = \frac 1x + o\left(\frac 1x\right), \quad (x\to +\infty) $$

The same can be accomplished in the second example:

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

Answer 2

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

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

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

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