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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 .

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2 Answers 2

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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

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  • $\begingroup$ You used taylor? also what are typos (I'm new here) $\endgroup$
    – El Bryan
    Dec 22, 2018 at 14:42
  • $\begingroup$ 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 $\endgroup$ Dec 22, 2018 at 14:47
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    $\begingroup$ Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence? $\endgroup$
    – El Bryan
    Dec 22, 2018 at 15:11
  • $\begingroup$ @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. $\endgroup$ Dec 22, 2018 at 15:15
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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: $$ \begin{align} \lim_{t\to 0} \frac{t - {t^3\over 3!} - t}{\sqrt{t} - {t\over 2} - \sqrt{t}} &= \lim_{t\to 0} \frac{{t^3\over 3!}}{{t\over 2}} = \\ &= \lim_{t\to 0} \frac{2t^3}{6t} = \\ &= \lim_{t\to 0} \frac{2t^2}{6} = 0 \end{align} $$

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