Since $\log(1+x) = x + O(x^2)$ when $x \to 0$, we have $n\log(1 + \frac{x}{n}) = x + O(\frac{x^2}{n})$ when $n\to +\infty$
Can someone explain the big-o notation here and why is it true?
I'm trying to understand a proof for the limit of $e^x$, but I really fail to understand the logic of the two equalities above.
For $\;|x|<1\;$:
$$\frac1{1+x}=1-x+x^2-x^3+\ldots\stackrel{\text{Integrate both sides elementwise}}\implies $$
$$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\ldots=x+\mathcal O(x^2)$$
For definitoni of big O, see here
We can show $\lim_{x \to 0}\dfrac{\log(1+x) - x}{x^2} = -\frac{1}{2}$ by several methods, using L'Hopital's rule for example.
This means for $|x|$ small enough, we have $1<\dfrac{\log(1+x) - x}{x^2} < 0$, i.e. $\left|\dfrac{\log(1+x) - x}{x^2}\right|$ is bounded. So by the definition of big O, $\log(1+x) - x = O(x^2)$ when $x \to 0$
We say $f(x)=\mathcal{O}(g(x))$ as $x\to a$ if there exist positive constants $\delta$ and $M$ such that $$|f(x)|\leq M|g(x)|$$ for $|x-a|<\delta$. (See http://en.wikipedia.org/wiki/Big_O_notation)
So, we should find constant $M$ such that $|\log (1+x)-x|\leq M x^2$ for sufficiently small $x$. We can show that specific $M$(ex. $M=\frac{1}{2}$) satisfies the inequality for small $x$ by applying mean value theorem for the function $(x-\log (1+x))-Mx^2$. (Note that $x-\log (1+x)\geq 0$ for $x\in (-1, 1)$)
This can be shown using Taylor series https://math.stackexchange.com/a/3413546/1028788:
$f(x) = f(0) + f'(0) \cdot x + O(x^{2})$, where
$f(x) = \log (1+x)$
$f(0) = \log (1+0) = 0$
$f'(0) = \frac{1}{1+0} = 1$
Thus we have $\log(1+x) = x + O(x^{2})$
Personally, I find using the equivalent small o notation more relevant here, since we are considering the limiting behaviour.
$\log(1+x) = x + o(x)$, as $x \to 0, o(x) \to 0$. ($o(x)$ represents a function that goes to zero faster than x)
Reference: An Introduction to Optimisation by Edwin Kah Pin Chong Page 72
