2
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ Are you allowed to use Taylor series expansions? $\endgroup$ – Daniel Schepler Jul 12 '17 at 22:53
  • $\begingroup$ @DanielSchepler Yes I am. $\endgroup$ – NapMaster Jul 12 '17 at 22:54
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Schepler Jul 12 '17 at 22:56
  • $\begingroup$ @DanielSchepler Thank you! $\endgroup$ – NapMaster Jul 12 '17 at 23:03
  • $\begingroup$ Just for the record, 0 is the limit as $x$ approaches 0 from the left. $\endgroup$ – MasterYoda Jul 12 '17 at 23:45
2
$\begingroup$

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 $\;\dfrac{\dfrac{x^2}2}{2x^2}=\dfrac14$.

$\endgroup$
7
  • $\begingroup$ Simple and clear approach. +1 $\endgroup$ – Paramanand Singh Jul 13 '17 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 '17 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 '17 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 '17 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 '17 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.