I have a problem with this limit. I have no idea where is the problem. Can you correct my mistake? Thanks

$$\lim\limits_{x \to 0} \left(\frac{x^5 e^\frac{-1}{x^2}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}\right)$$

I used the developments of McLaurin $e^x$ and $\sin x$

$$\lim\limits_{x \to 0} \left(\frac{x^5 (1-\frac{1}{x^2}+\frac{1}{2x^4})+\frac{x}{2} - ((\frac{x}{2})-(\frac{x^3}{48}))}{x^3}\right) = \lim\limits_{x \to 0} \left(\frac{x^5-x^3+ \frac{x}{2} +\frac{x}{2} - \frac{x}{2}+\frac{x^3}{48}}{x^3}\right)=$$

$$\lim\limits_{x \to 0} \left(\frac{-x^3+\frac{x^3}{48}}{x^3}\right)= \lim\limits_{x \to 0} \left(\frac{-\frac{47x^3}{48}}{x^3}\right)=\lim\limits_{x \to 0} \left(-\frac{47x^3}{48x^3}\right)= -\frac{47}{48}$$

but the result is wrong.

  • $\begingroup$ it should be $\frac{1}{48}$ $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '15 at 16:40
  • $\begingroup$ i know but howw $\endgroup$ – user12 Dec 29 '15 at 16:40
  • $\begingroup$ Is there any reason to believe that this limit could be evaluated purely by using L'Hopital? I suppose it would take a long time (you'd have to use the product rule many times), but could it be done? $\endgroup$ – zz20s Dec 29 '15 at 17:31

One may write, as $x \to 0$, $$ \begin{align} \frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}&=\frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \left(\frac{x}2-\frac{x^3}{48}+O(x^5)\right)}{x^3}\\\\ &=\frac{x^5 e^{-\frac1{x^2}} + \frac{x^3}{48}+O(x^5)}{x^3}\\\\ &=\frac{x^2 e^{-\frac1{x^2}} + \frac{1}{48}+O(x^2)}{1}\\\\ &= \frac{1}{48}+O(x^2)\\\\ \end{align} $$ giving $\dfrac{1}{48}$ for the sought limit, where we have used $\displaystyle x^2 e^{-\frac1{x^2}}=O(x^2)$ (to say the least).

  • $\begingroup$ so I dont need to develop the e^x ? $\endgroup$ – user12 Dec 29 '15 at 16:50
  • $\begingroup$ No -- even keeping it as an $o(1)$, you still have a remaining $x^2$ in front that guarantees that term can be discarded. Note that there is no "good" expansion of the exponential anyway -- its argument goes to $-\infty$. $\endgroup$ – Clement C. Dec 29 '15 at 16:52
  • $\begingroup$ @Amarildo You can not expand $\displaystyle e^{-\frac1{x^2}} $as you did, because the terms $\displaystyle O\left(\frac{1}{x^{2n}}\right)$ are not small as $x \to 0$. are you OK with this fact? $\endgroup$ – Olivier Oloa Dec 29 '15 at 16:52
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    $\begingroup$ ok i got it thanks $\endgroup$ – user12 Dec 29 '15 at 16:57

Setting $y=x^{-1}$, it is equivalent to $$ y^3 \cdot \left(\frac{1}{e^{y^2}y^5}+\frac{1}{2y}-\sin\left(\frac{1}{2y}\right)\right) \approx y^3 \cdot \left(\frac{1}{2y}-\left(\frac{1}{2y}-\frac{1}{48y^3}\right)\right) $$ as $y\to \infty$. Conclude.


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