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