# 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.

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\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}
If we let $t_n = \ln (1+ \frac{x}{\sqrt{n}})$, then $t_n \to 0$ as $n \to \infty$. Then we have $n = (\frac{x}{e^{t_n}-1})^2$, and the expression becomes $\frac{x^2}{e^{t_n}-1}- t \frac{x^2}{(e^{t_n}-1)^2} = x^2 \frac{(e^{t_n}-1-{t_n})}{(e^{t_n}-1)^2}$. Applying l'Hôpital's rule twice shows $\lim_{n \to \infty} \frac{(e^{t_n}-1-{t_n})}{(e^{t_n}-1)^2} = \frac{1}{2}$, from which the result follows.