How to find $\lim_{n\to\infty} \sqrt n \left(\log\left(\frac x{e\sqrt n} + 1\right)\right)$ As part of a larger problem (showing convergence in distribution of a random variable), I am trying to find $\displaystyle{\lim_{n\to\infty}} \sqrt n \left(\log\left(\frac x{e\sqrt n} + 1\right)\right)$. I know that the answer is $\frac xe$ (via WolframAlpha) but I don't know how to show it.
So far I have tried manipulating the expression to make it amenable to L'Hopital's rule, but this hasn't succeeded; I wind up just getting another indeterminate expression after applying L'Hopital's. Any tips would be appreciated.
 A: The systematic, "let's not think about tricks and do Taylor series" way:
As $\frac{x}{e\sqrt{n}}\xrightarrow[n\to\infty]{} 0$ for every fixed $x$, we have
$$
\log\left(1+\frac{x}{e\sqrt{n}}\right) = \frac{x}{e\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right)
$$
from which
$$
\sqrt{n}\log\left(1+\frac{x}{e\sqrt{n}}\right) = \frac{x}{e} + o\left(1\right)\xrightarrow[n\to\infty]{}\frac{x}{e}
$$
where we used the first-order Taylor series for $\log(1+u)$ around $0$.

The more elementary, simpler way (once you recognize a derivative): use
$$
\sqrt{n}\log\left(1+\frac{x}{e\sqrt{n}}\right) = \frac{x}{e}\cdot \frac{\log\left(1+\frac{x}{e\sqrt{n}}\right)}{\frac{x}{e\sqrt{n}}} \xrightarrow[n\to\infty]{} \frac{x}{e}\cdot\frac{d}{du} \log(1+u)\big|_{u=0} = \frac{x}{e}\cdot1
$$
Note that this is equivalent to the first method, since the derivative here amounts to first-order Taylor approximation.
A: I know you have an answer, but I think there's an easier way.
Let $a = x/e$ and $y = \sqrt n$
$$\lim_{\sqrt n \to \infty} \sqrt n \ln (\frac{x}{e\sqrt n} + 1) = \ln (\lim_{y \to \infty} (\frac{a}{y}+ 1)^y) $$
The latter is a common limit problem, which gives us
$$\ln (\lim_{y \to \infty} (\frac{a}{y}+ 1)^y) = \ln e^a = a = \frac{x}{e}$$
A: 
I thought it might be instructive to present an approach that relies on elementary, pre-calculus tools only.  To that end, we now proceed.


In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$
for all $x>0$.
Using $(1)$ with $x$ replaced with $1+\frac{x}{e\sqrt n}$ reveals
$$\frac{\frac{x}{e}}{1+\frac{x}{e\sqrt n}}\le \sqrt{n}\log\left(1+\frac{x}{e\sqrt n}\right)\le \frac{x}{e}\tag2$$
whence applying the squeeze theorem to $(2)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\sqrt{n}\log\left(1+\frac{x}{e\sqrt n}\right)=\frac xe}$$
A: Hint:  $\lim_{n\to\infty} \sqrt n \left(\log\left(\frac x{e\sqrt n} + 1\right)\right)=\lim_{n\to\infty}\frac{  \left(\log\left(\frac x{e\sqrt n} + 1\right)\right)}{\frac{1}{\sqrt n}}$, now use L'Hopital's rule.
