Is this a justifiable step in $\lim_{x\to \infty}(1+\sin(1/x))^x$ So I have $\lim_\limits{{x\to \infty}}(1+\sin\frac{1}{x})^x$. I believe that this equals $e$. I know that $0 < \lim_\limits{{x\to \infty}}\sin\frac{1}{x}< \lim_\limits{{x\to \infty}}\frac{1}{x} = 0$. Can I just replace $\sin\frac{1}{x}$ with $\frac{1}{x}$? If so, then I get $\lim_\limits{{x\to \infty}}(1+\frac{1}{x})^x$, which is the definition of $e$.
 A: $$\lim_{x\to \infty}\left(1+\sin\frac{1}{x}\right)^x = \lim_{z\to 0}(1+\sin z)^{1/z} = \exp\lim_{z\to 0}\frac{\log(1+\sin z)}{z}\stackrel{d.H.}{=}\exp\lim_{z\to 0}\frac{\cos z}{1+\sin z} $$
clearly equals $\exp 1 = \color{red}{e}$.
A: The answer does turn out to be $e$, but I think it would be more work to justify replacing $\sin \frac 1x$ by $\frac 1x$, then to just work out the limit in the usual way.  Take the limit of the log instead and use L'hopital.  The answer pops right out.
A: Calculate the limit of the log, and use equivalents:
$$\ln\biggl(1+\sin\frac1x\biggr)^{\!x}=\frac1x\ln\biggl(1+\sin\frac1x\biggr).$$


*

*$\ln(1+u)\sim_0u$,

*$\sin\dfrac1x\sim_\infty \dfrac1x$,
hence
$$\ln\biggl(1+\sin\frac1x\biggr)^{\!x}\sim_\infty x\ln\biggl(1+\frac1x\biggr)\sim_\infty x\cdot\frac1x=1.$$

A: It is a simple exercise to check the following theorem.

Let $u(x)$ and $\omega(x)$ be functions of one variable such that 
  $$
u(x)>\omega(x)\qquad \mbox{ and }\qquad\lim_{x\to \infty} \left [u(x)-\omega(x)\right ] =0.
$$
  Let $f(x,y)$ be a function of two variables such that
  $$
\lim_{x\to \infty}f(x,u(x))=L\qquad
\lim_{x\to \infty}f(x,\omega(x))=M
$$
  If the partial derivatives $D_1f(x,y)$ and $D_2f(x,y)$ are continuous and  $\lim_{x\to\infty}D_2f(x,y(x))=K$ for $\omega(x)\leq y(x)\leq u(x)$ then 
  $$
\lim_{x\to \infty}f(x,\omega(x))=L
$$

Proof. We have 
$$
\lim_{x\to \infty}f(x,u(x))
=\lim_{x\to \infty}f(x,u(x))-f(x,\omega(x))+f(x,\omega(x)).
$$
By Mean value theorem in several variables we have 
\begin{align}
\lim_{x\to \infty}f(x,u(x))
=& \lim_{x\to \infty}D_2 f(x,y(x))\cdot [u(x)-\omega(x)]
+ \lim_{x\to \infty}f(x,\omega(x))
\\
=& \lim_{x\to \infty}D_2 f(x,y(x))\cdot \lim_{x\to \infty}[u(x)-\omega(x)]
+ \lim_{x\to \infty}f(x,\omega(x))
\\
=&\lim_{x\to \infty}f(x,\omega(x))
\\
\end{align}
Set $f(x,y)=(1+y)^x$, $u(x)=1/x$ and $\omega(x)=(1/x)\frac{\sin(1/x)}{(1/x)}$ and good luck.
A: $$\log L=x\log\left(1+\sin\left(\frac{1}{x}\right)\right) \\ x\log(1+1/x+O(x^{-3}))=\\x(1/x+O(x^{-2})) =\\ 1+O(x^{-1})\\
\stackrel{x\to\infty}{\to}1$$
Thus, $L=e$.
A: Do remember that in mathematics we can't replace $A$ by $B$ unless $A=B$. So replacing $\sin(1/x)$ with $1/x$ is not allowed. What is however allowed is that you can replace $\lim_{x\to\infty} \sin(1/x)$ with $\lim_{x\to\infty} 1/x$ as both are equal to $0$.
The limit in question is not that difficult to evaluate. We put $x=1/t$ so that $t\to 0^{+}$ and all the limits below will be based on this context. The desired limit is then equal to
\begin{align}
L&=\lim(1+\sin t) ^{1/t}\notag\\
&=\lim\exp\log(1+\sin t) ^{1/t}\notag\\
&=\lim\exp\left(\frac{\log(1+\sin t)} {t} \right)\notag\\
&=\exp\lim\frac{\log(1+\sin t)} {\sin t} \cdot\frac{\sin t} {t}\notag \\
&=\exp(1\cdot 1)\notag\\
&=e\notag
\end{align} 
