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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.

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    $\begingroup$ $$\lim_{x\to \infty} \dfrac{\log(1/x + 1)}{1/x} = 1$$. $\endgroup$
    – user312097
    Sep 12, 2017 at 14:58
  • $\begingroup$ Maybe use some log properties to break up that $\frac{x}{e\sqrt{n}}$ into subtraction? Possibly combine the 1 and the fraction into a single term then break up into separate logarithms. Good luck! $\endgroup$
    – Jon Staggs
    Sep 12, 2017 at 15:01
  • $\begingroup$ Set it equal to $L$ and then raise both sides to $e^L=exp(\sqrt{n}\ln(\frac{x}{e\sqrt{n}}+1)$, then solve. $\endgroup$
    – JohnColtraneisJC
    Sep 12, 2017 at 15:01
  • $\begingroup$ So many different ways to solve this. I would use the taylor expansion of $\log(1+x)$ around zero. $\endgroup$
    – Marc
    Sep 12, 2017 at 15:05

4 Answers 4

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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

$$\ln (\lim_{y \to \infty} (\frac{a}{y}+ 1)^y) = \ln e^a = a = \frac{x}{e}$$

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  • $\begingroup$ With this answer I finally feel that I am able to understand how to reach the solution (rather than merely follow along another's reasoning). Thank you. $\endgroup$
    – Ceph
    Sep 12, 2017 at 17:22
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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.

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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

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\sqrt{n}\log\left(1+\frac{x}{e\sqrt n}\right)=\frac xe}$$

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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.

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  • $\begingroup$ So... you're using L'Hopital rule, a (not-so-elementary) theorem which involves computing $f'(u)$ and $g'(u)$ to compute the limit of $\frac{f(u)}{g(u)}$ in order to find the limit $\lim_{u\to 0}\frac{f(u)-f(0)}{u-0}$? Isn't that a bit... unnecessary, and deeply circular? $\endgroup$
    – Clement C.
    Sep 12, 2017 at 15:12
  • $\begingroup$ @ClementC. I agree to your comment. $\endgroup$
    – Arun
    Sep 13, 2017 at 4:09

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