Find $\lim\limits_{n\rightarrow \infty} n\ln\left(1+\frac{x}{\sqrt{n}}\right)-x\sqrt{n}$ As the title suggests, I need to prove the $$\lim_{n\rightarrow \infty}  n\ln\left(1+\frac{x}{\sqrt{n}}\right)-x\sqrt{n} = \frac{-x^2}{2}$$
I've tried to used things like L'hospitals but to no avail. Any ideas?

Let's put $t = 1/\sqrt{n}$ so that $n = 1/t^{2}$ and then we have
\begin{align}
L &= \lim_{n \to \infty}n\log\left(1 + \frac{x}{\sqrt{n}}\right) - x\sqrt{n}\notag\\
&= \lim_{t \to 0}\frac{1}{t^{2}}\log\left(1 + tx\right) - \frac{x}{t}\notag\\
&= \lim_{t \to 0}\frac{\log(1 + tx) - tx}{t^{2}}\notag\\
&= \lim_{t \to 0}\dfrac{\dfrac{x}{1 + tx} - x}{2t}\text{ (via L'Hospital's Rule)}\notag\\
&= -\frac{1}{2}\lim_{t \to 0}\frac{x^{2}}{1 + tx}\notag\\
&= -\frac{x^{2}}{2}\notag
\end{align}
Thus L'Hospital's Rule works fine here.
