I want to use Taylor's Theorem to show that $\ln(1+\frac{x}{n}) = \frac{x}{n}+O\frac{x^2}{n^2}$ (I cannot use $\ln(1+x) = \sum_{n=1}^{\infty}(-1)^{n}\frac{x^n}{n}$ in this question).
My attempt:
Apply Taylor's Theorem, for each pair of points $x,x_0 \in (a,b)$, $\exists c \in (x,x_0)$ s.t. $$ \ln(1+\frac{x}{n}) = f(x_0)+f'(x_0)(x-x_0)+\frac{f''(c)}{2!}(x-x_0)^2 \\ = \ln(1+\frac{x_0}{n})+\frac{x-x_0}{n+x_0}-\frac{(x-x_0)^2}{2!(n+c)^2} $$ This does not seem to align with $ \frac{x}{n}+O\frac{x^2}{n^2}$, any idea on how to approach this?
You're very close. Put $x_0 = 0$ and $1/(n+c)^2 = O(1/n^2) $.