<s>[INST] Evaluating $\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^3\sin x}$ 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.
 [/INST] The following steps use De l'Hopital rule.
$$\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)}\right)=\lim _{x\to \:0}\left(\frac{-e^{-\frac{x^2}{2}}x+\sin \left(x\right)}{3x^2\sin \left(x\right)+\cos \left(x\right)x^3}\right)=\lim _{x\to \:0}\left(\frac{e^{-\frac{x^2}{2}}x^2-e^{-\frac{x^2}{2}}+\cos \left(x\right)}{-x^3\sin \left(x\right)+6x^2\cos \left(x\right)+6x\sin \left(x\right)}\right)$$
$$=\lim _{x\to  0}\left(\frac{-e^{-\frac{x^2}{2}}x^3+3e^{-\frac{x^2}{2}}x-\sin \left(x\right)}{-x^3\cos \left(x\right)-9x^2\sin \left(x\right)+18x\cos \left(x\right)+6\sin \left(x\right)}\right)=\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}x^4-6e^{-\frac{x^2}{2}}x^2+3e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)-12x^2\cos \left(x\right)-36x\sin \left(x\right)+24\cos \left(x\right)}\right)$$
$$=\frac{e^{-\frac{0^2}{2}}\cdot  0^4-6e^{-\frac{0^2}{2}}\cdot  0^2+3e^{-\frac{0^2}{2}}-\cos \left(0\right)}{0^3\sin \left(0\right)-12\cdot  0^2\cos \left(0\right)-36\cdot  0\cdot \sin \left(0\right)+24\cos \left(0\right)}=\frac 1{12}. \square$$
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