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$$\lim_{x\to\infty}\left(\frac{({e^{1/x}-e^{\sin(1/x)})}{x^3}}{{(x!)}^{(1/x)}}\right)$$

I have been trying to find the limit of the following sequence above but I am stuck. I first tried using substitution for $\sin(1/x)$ because it approaches $0$; however, that would make the entire expression $0$ which is obviously wrong.

I next tried using the laws of logs, but that just made it more complicated getting nowhere.

zhw.
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Math-Man
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    Is x an non-negative integer? If not then the factorial is not defined, but is defined if you use the Gamma function. – Andrew Shedlock Jan 03 '19 at 01:26
  • The denominator I think is asymptotic to $x/e$ (for example using Stirling's asymptotic formula), but numerator seems to be asymtptotic to $1/6$, which would make the limit $0$. If I replace $x^3$ with $x^4$ at the numerator then, I get $e/6$ as limit... Does it make any sense to you? If it does I can show you details of my calculations. – dfnu Jan 03 '19 at 01:34
  • We haven't learned the Gamma function, yet. Thank you though. – Math-Man Jan 03 '19 at 02:56

2 Answers2

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First note that $x/(x!) ^{1/x}\to e$ where by $x! $ we mean $\Gamma (x+1)$ and hence the desired limit is equal to the limit of the expression $$e\cdot(e^{1/x}-e^{\sin(1/x)})x^2$$ Putting $x=1/t$ so that $t\to 0^{+}$ we see that the desired limit is $$e\lim_{t\to 0^{+}}e^{\sin t} \cdot\frac{e^{t-\sin t} - 1}{t-\sin t} \cdot \frac{t-\sin t} {t^2}$$ Clearly the first two factors under limit operation tend to $1$ and the last one tends to $0$ and thus the limit in question is $0$.

The fact that $x/\Gamma ^{1/x}(x+1)$ tends to $e$ as $x\to\infty $ can be established in the following manner.

Let $f(x) =\log(x^x/\Gamma (x+1))$ then $$f(x+1)-f(x)=x\log\left(1+\frac {1}{x}\right) $$ which tends to $1$ and hence $f(x) /x\to 1$. And thus $\log (x/\Gamma ^{1/x}(x+1))\to 1$.

The case when $x$ is an integer is famous and we can prove that $\lim\limits _{n\to\infty} \dfrac{n} {\sqrt[n] {n!}} =e$ using the well known

Theorem: If $\{a_n\} $ is a sequence of positive terms such that $a_{n+1}/a_n\to L$ as $n\to\infty $ then $\sqrt[n] {a_n} \to L$ as $n\to\infty $.

Taking $a_n=n^n/n! $ we can see that $$\frac{a_{n+1}}{a_n}=\left(1+\frac {1}{n}\right) ^n\to e$$ so that $\sqrt[n] {a_n} =n/\sqrt[n] {n!} \to e$.

A similar theorem like the one above holds for functions of a real variable and the same has been used in my solution which uses $x$ as a real variable.

Paramanand Singh
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  • Thank you for your solution. I am curious of how did you know it tends to e and what gave you the inclination to use substitution? – Math-Man Jan 03 '19 at 06:58
  • @DavidMB: the limit $\lim_{n\to\infty} n/\sqrt[n] {n!} =e$ ($n$ integer) is a famous exercise and is already available on this website. I have given a similar argument replacing integer $n$ with real variable $x$. – Paramanand Singh Jan 03 '19 at 09:07
  • @DavidMB: I have added an update in my answer with some details. – Paramanand Singh Jan 03 '19 at 09:22
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\begin{align} &\lim_{x \to \infty}\frac{(\exp(1/x) - \exp(\sin(1/x)))x^3}{(x!)^\frac1x} \\&= \lim_{x \to \infty}\frac{(1+\frac1x+ \frac1{2x^2} +\frac1{6x^3}- \exp(\frac1x-\frac1{6x^3}))x^3}{(x!)^\frac1x}\\ &=\lim_{x \to \infty}\frac{(1+\frac1x+ \frac1{2x^2}+\frac1{6x^3} - \left(1+\frac1x+\frac1{2x^2} -\frac1{6x^3}+\frac1{6x^3}\right))x^3)}{(x!)^\frac1x}\\ &=\frac16 \cdot \lim_{x \to \infty} \frac1{\exp\left( \frac{\ln x!}{x}\right)}\\ &= 0 \end{align}

Siong Thye Goh
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