Prove that if $n$ is a positive integer and $|x| \leq \dfrac{1}{2}n$
then $(i)\quad n\log\left(1+\dfrac{x}{n}\right)=x+Q_{n}(x)$
where $(ii)\quad |Q_{n}(x)|\leq\dfrac{|x|^{2}}{n}$
and deduce that $(iii)\quad \left(1+\dfrac{x}{n}\right)^n \rightarrow \exp(x)$ as $n \rightarrow \infty$
Now my attempt at the solution;
Using taylor series about 0; we get
$$f(x)=n\log\left(1+\dfrac{0}{n}\right)+\dfrac{1}{\left(1+\dfrac{0}{n}\right)}(x-0)+Q_{n}(x)$$
$$f(x)=0+x+Q_{n}(x)$$ so that answers part (i)
For part (ii) we take the remainder theorem which I think in this case should be
$$Q_{n}(x)=\dfrac{(x-0)^2}{(2+1)!}f^{3}(0)$$
Is this correct , I dont think so , any help to continue the question would be much appreciated.
$2)$ Using the formulae for remainder $$Q_{n}(x) = \frac{x^2}{2!} \cdot \frac{-n}{(n+\xi)^2} \le 0 \le \frac{x^2}{ 2 n}, \mathrm{ where }, \xi\in \left( -\frac 1 2 n, \frac 1 2 n\right)$$
$3)$ $$\lim_{n\to\infty}\quad n\log\left(1+\dfrac{x}{n}\right)= \lim_{n\to\infty}x+ \frac{|x|^2}{n} = x \\ \implies \lim_{n\to\infty}\left( 1 + \frac x n \right)^n = e^x$$
A bit elaboration of $3)$.
Using property of log $ a \log x = \log x^a$, $$\implies \lim_{n\to\infty} \log \left( \left(1 + \frac x n \right )^n \right) = \log \left( \lim_{n\to\infty} \left(1 + \frac x n \right )^n \right) - x$$
Since $e$ and $\log$ are inverse of each other, we take $e^{\mathrm{on \, both\, sides}}$ and get the result.