<s>[INST] Show $x\sqrt{n} - n \ln\left(1+\frac{x}{\sqrt{n}}\right) \to \frac{x^2}{2}$ With $x > 0$, show
$$
L=\lim_{n \to \infty} x\sqrt{n} - n \ln\left(1+\frac{x}{\sqrt{n}}\right) = \frac{x^2}{2}. 
$$
I tried to write
$$
x\sqrt{n}=\ln \left( e^{x\sqrt{n}} \right),
$$
so that 
$$
L  = \lim_{n\to \infty}
\frac{e^{x\sqrt{n}}}{\left(1+x/\sqrt{n}\right)^n}.
$$
The last expression is of the form $\frac{\infty}{\infty}$.
However, l'hospital rule won't change the denumerator.  
I don't see what to do.  
 [/INST] $$\begin{aligned}x\sqrt{n}-n\ln\left(1+\frac{x}{\sqrt{n}}\right)&=x\sqrt{n}-n\left(\frac{x}{\sqrt{n}}-\frac{x^2}{2n}+\frac{x^3}{3n\sqrt{n}}-\cdots\right)\\&=\frac{x^2}{2}-\frac{x^3}{3\sqrt{n}}+\cdots\to\frac{x^2}{2}\end{aligned}$$
</s>