# 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 .

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

• You used taylor? also what are typos (I'm new here) Dec 22, 2018 at 14:42
• I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit Dec 22, 2018 at 14:47
• Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence? Dec 22, 2018 at 15:11
• @ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity. Dec 22, 2018 at 15:15

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}