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I am trying to evaluate the following limit $$\lim_{x\to 0}\left\{\frac{e^{-x^2}\sin|x|-|x|}{|x^3|}\right\}$$

I have tried by using L'Hôpital's rule, but the expression becomes very messy after successive differentiation. I also do not know how to approach a limit involving modulus (not being possible to write this as a piece wise function).

Thanks in advance.

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  • $\begingroup$ Try finding the limit with the added restriction $x \ge 0$. $\endgroup$ – copper.hat Jun 14 '18 at 17:39
  • $\begingroup$ Do you know Taylor expansions? $\endgroup$ – mathcounterexamples.net Jun 14 '18 at 17:42
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We should be looking for $a_3$ where$$e^{-x^2}\sin x=\sum_{n=0}^{\infty}a_nx^n$$also we know that$$e^{-x^2}=1-x^2+\dfrac{x^4}{2}-\cdots\\\sin x=x-\dfrac{x^3}{6}+\cdots$$therefore$$e^{-x^2}\sin x=x-\dfrac{7}{6}x^3+o(x^3)$$therefore$$\lim_{x\to 0}\frac{e^{-x^2}\sin|x|-|x|}{|x^3|}=\lim_{|x|\to 0}\frac{e^{-x^2}\sin|x|-|x|}{|x^3|}=\lim_{x\to 0}\frac{e^{-x^2}\sin x-x}{x^3}=\lim_{x\to 0}\frac{x-\dfrac{7}{6}x^3+o(x^3)-x}{x^3}=-\dfrac{7}{6}$$

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