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I'm trying to evaluate $$\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^3\sin x}.$$

I can see the $\frac{0}{0}$ form, so I'll use L'Hôpital's rule. However, I'll eliminate the sine function in the denominator by multiplying the numerator and denominator by $x$. We know

$$\lim\limits_{x\rightarrow0} \frac{x}{\sin x} = 1.$$

The problem reduces to

$$\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^4}.$$

Now I'll use L'Hôpital's rule. The problem now becomes

$$\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}(-x)+\sin x}{4x^3}.$$

I don't know how to proceed from here onwards. Using L'Hôpital's rule any more complicates the problem. Any ideas would be appreciated.

Integreek
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StrawberryVanilla
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  • Are you allowed to use https://en.wikipedia.org/wiki/Taylor_series? – zkutch Feb 06 '23 at 12:24
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    Unpopular(?) opinion: L'Hôpital's rule is a stupid rule. – nejimban Feb 06 '23 at 12:25
  • @zkutch Yes, I'd like to know how to solve it that way. – StrawberryVanilla Feb 06 '23 at 12:27
  • @nejimban I'd appreciate learning any other method for solving such problems. – StrawberryVanilla Feb 06 '23 at 12:29
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    Using Taylor series is how I would prove L'Hôpital's rule and is much more robust. See: $\mathrm e^{-\frac{x^2}2}=1-\frac{x^2}2+\frac{x^4}8+o(x^4)$ and $\cos(x)=1-\frac{x^2}2+\frac{x^4}{24}+o(x^4)$ gives you $\mathrm e^{-\frac{x^2}2}-\cos(x)\sim\frac{x^4}{12}$. Now $\sin(x)\sim x$ so $$\frac{\mathrm e^{-\frac{x^2}2}-\cos(x)}{x^3\sin(x)}\sim\frac{\frac{x^4}{12}}{x^3\cdot x}=\frac1{12}.$$ – nejimban Feb 06 '23 at 12:32
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    @nejimban I wish I could upvote your comments more than once. – TheSilverDoe Feb 06 '23 at 12:37
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    @nejimban Don't know why, but I also have an aversion to L'H rule. It's overrated – Sine of the Time Feb 06 '23 at 20:19
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    @nejimban: I totally agree. There is nothing you can do with L'Hopital that cannot be done better using asymptotic expansion, as demonstrated in the posts linked from my profile. – user21820 Feb 07 '23 at 04:12
  • Note that you should say that your asymptotic statements are under the limiting condition of "as $x→0$". – user21820 Feb 07 '23 at 04:14
  • @SineoftheTime: You might also be interested to look at the posts I mentioned, as well as this. – user21820 Feb 07 '23 at 04:18
  • @user21820 Interesting reads, thank you! Even though L'Hôpital was a French mathematician, I find it amusing that his rule is generally ignored in French schools and universities (we prefer to emphasize on asymptotic expansions and Taylor series) but remains quite popular abroad. – nejimban Feb 07 '23 at 06:06
  • @nejimban: Hahaha I didn't know that! Thanks for the interesting note! – user21820 Feb 07 '23 at 06:19
  • Those who devalorize L'Hospital's rule forget that the Taylor expansion is obtained via several differentiations of the function(s). Hence there is no essential difference. –  Oct 21 '24 at 12:14

3 Answers3

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An elegant way to solve this problem, as suggested by @nejimban, is to use the Taylor series.

$\mathrm e^{-\frac{x^2}2}$ can be written as $$1-\frac{x^2}2+\frac{x^4}8+o(x^4)$$

whereas $\cos(x)$ can be written as $$1-\frac{x^2}2+\frac{x^4}{24}+o(x^4)$$

Now on subtracting,

$$\mathrm e^{-\frac{x^2}2}-\cos(x)\sim\frac{x^4}{12} \;\;\text{ and }\;\; \sin(x)\sim x$$

On substituting these values in the original limit-

$$\frac{\mathrm e^{-\frac{x^2}2}-\cos(x)}{x^3\sin(x)}\sim\frac{\frac{x^4}{12}}{x^3\cdot x}=\frac1{12}.$$

Gary
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StrawberryVanilla
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The following steps use L'Hôpital's rule.

$$\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)}\right)=\lim _{x\to \:0}\left(\frac{-e^{-\frac{x^2}{2}}x+\sin \left(x\right)}{3x^2\sin \left(x\right)+\cos \left(x\right)x^3}\right)=\lim _{x\to \:0}\left(\frac{e^{-\frac{x^2}{2}}x^2-e^{-\frac{x^2}{2}}+\cos \left(x\right)}{-x^3\sin \left(x\right)+6x^2\cos \left(x\right)+6x\sin \left(x\right)}\right)$$ $$=\lim _{x\to 0}\left(\frac{-e^{-\frac{x^2}{2}}x^3+3e^{-\frac{x^2}{2}}x-\sin \left(x\right)}{-x^3\cos \left(x\right)-9x^2\sin \left(x\right)+18x\cos \left(x\right)+6\sin \left(x\right)}\right)=\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}x^4-6e^{-\frac{x^2}{2}}x^2+3e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)-12x^2\cos \left(x\right)-36x\sin \left(x\right)+24\cos \left(x\right)}\right)$$ $$=\frac{e^{-\frac{0^2}{2}}\cdot 0^4-6e^{-\frac{0^2}{2}}\cdot 0^2+3e^{-\frac{0^2}{2}}-\cos \left(0\right)}{0^3\sin \left(0\right)-12\cdot 0^2\cos \left(0\right)-36\cdot 0\cdot \sin \left(0\right)+24\cos \left(0\right)}=\boxed{\frac 1{12}}$$

Integreek
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Sebastiano
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    You would have avoided all the complexity in the denominator by substituting $x$ for $\sin(x)$. –  Oct 21 '24 at 12:11
  • @YvesDaoust You're right. I would't change the answer. :-( After your suggestion I see your hint. – Sebastiano Oct 21 '24 at 16:03
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We have that

$$\frac{e^{-\frac{x^2}{2}}-\cos x}{x^3\sin x}=\frac{x}{\sin x}\left(\frac14\frac{e^{-\frac{x^2}{2}}-1+\frac{x^2}{2}}{\frac{x^4}{4}}+\frac{1-\frac{x^2}{2}-\cos x}{x^4}\right) \to 1\cdot \left(\frac14\cdot\frac12-\frac1{24}\right)=\frac1{12}$$

indeed, as shown here, as $x \to 0$ we have that $\frac{e^x-1-x}{x}\to \frac12$ and $\frac{\sin x -x}{x^3}\to -\frac16$, therefore by $x=2y$

$$\frac{1-\frac{x^2}{2}-\cos x}{x^4}=\frac{\sin^ 2y-y^2}{8y^4}=\frac18\frac{\sin y + y}{y}\cdot\frac{\sin y - y}{y^3}=\frac18 \cdot2\cdot \left(-\frac16\right)=-\frac1{24}$$

user
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