Help understanding the steps of a solved limit A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:
\begin{align}
\lim_{x \rightarrow +∞}
\frac{\sin (1/x) - (1/x)}{\log(1+(1/\sqrt{x}))-\sqrt{1/x}}
&=\lim_{x \rightarrow +∞}
\frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
\\
&=
\lim_{x \rightarrow +∞}
\frac{2x}{(6x^3)}= 0
\end{align}
The part that I don't understand is why the limit is equal to:
\begin{equation*}
\frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
\end{equation*}
The only thing I found out so far is that:
\begin{equation*}
\sin (x) - x ∼ x^3/6
\end{equation*}
So:
\begin{equation*}
\sin (1/x) - (1/x)∼ 1/(6x^3)
\end{equation*}
For the rest, I have no idea .
 A: It isn't exactly correct. 
$$\sin(x)=x-x^3/3!+x^5/5!-\cdots$$
This can be obtained from Taylor's expansion.
For $\log(1+x),$
$$\log(1+x)=\int\frac{1}{1+x}dx$$
$$=\int (1-x+x^2-x^3+\cdots)dx$$
$$\log(1+x)=x-x^2/2+\cdots$$
So, your limit,
\begin{equation*}
\lim_{x \rightarrow +∞}
\frac{\sin (1/x) - (1/x)}{\log(1+(1/\sqrt{x}))-(1/\sqrt x)}=
\end{equation*}
\begin{equation*}
\lim_{x \rightarrow +∞}
\frac{(1/x) -(1/6x^3)\cdots - (1/x)}{(1/\sqrt x)-(1/2x)\cdots-(1/\sqrt x)}=
\end{equation*}
Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
\begin{equation*}
\lim_{x \rightarrow +∞}
\frac{2x}{6x^3}= 0
\end{equation*}
Your solution has a lot of typos
A: First note that $x\to \infty$, perform substitution $t = {1\over x}$, then $t\to 0$:
$$
\lim_{t\to 0} \frac{\sin t - t}{\log(1 + \sqrt{t}) - \sqrt{t}}
$$
By Taylor expansion you may approximate $\sin t$ at  $t = 0$ by:
$$
\sin t = t - {t^3\over 3!} + {t^5\over 5!} - \dots
$$
At the same time for $\log(1+t)$:
$$
\log(1+t) = t - {t^2\over 2} + {t^3\over 3}-\cdots
$$
So if you apply this to your limit you'll observe the desired result:
$$
\begin{align}
\lim_{t\to 0} \frac{t - {t^3\over 3!} - t}{\sqrt{t} - {t\over 2} - \sqrt{t}} &= \lim_{t\to 0} \frac{{t^3\over 3!}}{{t\over 2}} = \\ 
&= \lim_{t\to 0} \frac{2t^3}{6t} = \\
&= \lim_{t\to 0} \frac{2t^2}{6} = 0
\end{align}
$$
