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I'm trying to solve the following limit:

$$\lim_{x\rightarrow 0^+} \frac{\displaystyle\sqrt{x^2+x^3} - \sin(x)}{\displaystyle 2x^2 - {e}^{-1/x}}$$

For WolframAlpha the result is: $ \frac14 $, while, according to my calculations, it is: $0$.

The text forbids me to use L'Hôpital's rule. Is the answer given by WolframAlpha wrong? or am I?

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  • $\begingroup$ Are you allowed to use Taylor series expansions? $\endgroup$ Jul 12, 2017 at 22:53
  • $\begingroup$ @DanielSchepler Yes I am. $\endgroup$
    – NapMaster
    Jul 12, 2017 at 22:54
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    $\begingroup$ Then I'd first get rid of the $e^{-1/x}$ term using $\lim_{x\to 0^+} \frac{2x^2}{2x^2 - e^{-1/x}} = 1$ (which follows from $\lim_{x \to 0^+} \frac{e^{-1/x}}{x^2} = 0$). Then expand what remains on top and bottom as Taylor series. $\endgroup$ Jul 12, 2017 at 22:56
  • $\begingroup$ @DanielSchepler Thank you! $\endgroup$
    – NapMaster
    Jul 12, 2017 at 23:03
  • $\begingroup$ Just for the record, 0 is the limit as $x$ approaches 0 from the left. $\endgroup$
    – MasterYoda
    Jul 12, 2017 at 23:45

1 Answer 1

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I'll systematise using equivalents:

Since $\mathrm e^{-1/x}=o\bigl(x^2\bigr)$, we know $2x^2+\mathrm e^{-1/x}\sim_0 2x^2$.

Let's expand the numerator with Taylor's formula at order $2$: $$\sqrt{x^2+x^3}-\sin x=x\sqrt{1+x}-\sin x=x\Bigl(1+\frac x2\Bigr)-x+o\bigl(x^2\bigr)=\frac{x^2}2+o\bigl(x^2\bigr)\sim_0\frac{x^2}2,$$ so the function is equivalent, near $0$, to $\;\dfrac{\dfrac{x^2}2}{2x^2}=\dfrac14$.

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  • $\begingroup$ Simple and clear approach. +1 $\endgroup$
    – Paramanand Singh
    Jul 13, 2017 at 5:16
  • $\begingroup$ I was doing wrong ${e}^{-1/x}$. I was not considering that the Taylor series with center 0 (which is the Maclaurin series) is identically zero. But what does change when $x$ approaches to $0$ from the right? $\endgroup$
    – NapMaster
    Jul 13, 2017 at 10:31
  • $\begingroup$ I'm not sure to fully understand what you mean, but I'd say the Taylor's series for $\mathrm e^{-1/x}$ is irrelevant here, as the function is not the sum of its Taylor's series. Instead we need a more general ‘comparison scale’, which I used under the form $\mathrm e^{-1/x}=o(x^2)$, in order to obtain an equivalent of the denominator. $\endgroup$
    – Bernard
    Jul 13, 2017 at 10:38
  • $\begingroup$ @Bernard so, you just derived two times ${e}^{-1/x}$ and you discovered that it is always $0$ ? I'm trying to understand for which reason you said that ${e}^{-1/x} = o(x^2)$. My other question was: How does the result change when $x$ approaches to $0$ form the right? $\endgroup$
    – NapMaster
    Jul 13, 2017 at 10:56
  • $\begingroup$ I didn't derived only twice. It's a well-known exercise that such functions, continued at $0$ have (right-sided) derivatives of all orders equal to $0$ at $0$. For the next question, set $u=1/x$. You have $\dfrac{\mathrm e^{-1/x}}{x^2}=\dfrac{u^2}{\mathrm e^u}\to 0$ when $u\to+\infty$.. $\endgroup$
    – Bernard
    Jul 13, 2017 at 11:08

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