Calculate the value of the following limit So I have given the limit:
$$\lim_{x\to0} \frac{2\sin\left ( e^{-\frac{x^2}{2}} -\cos x \right)}{(\arctan(\sinh(x^2))^2}$$
I have been struggling for hours with it. Since i got the undefined form when i put $x=0$ i tried out with L'Hopital method and I come to this point:
$$\lim_{x\to 0} \frac{2\cos(e^{-\frac{x^2}{2}}-\cos x)(e^{-\frac{x^2}{2}}(-x)+\sin x)}{\frac{4x \arctan(\sinh(x^2))\cosh(x^2)}{\sinh(x^2)+1}}$$
Still here when i substitute x with $0$ i still get $0$. I tried factorising the x, i also tried using the identities $cos(A-B)$ and so on, but nothing.
I think the answer that should come out is $\frac{1}{6}$
I would be very thankful for your help,
Annalisa

We have that
$$\frac{2\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{(\arctan(\sinh(x^2))^2}
=$$
$$=\left(\frac{\sinh(x^2)}{\arctan(\sinh(x^2)}\right)^2\cdot\left(\frac{x^2}{\sinh(x^2)}\right)^2\cdot \frac{\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{e^{\frac{-x^2}2}-\cos x }\cdot2\cdot\frac{e^{\frac{-x^2}2}-\cos x }{x^4} $$
and since by standard limits
$$\left(\frac{\sinh(x^2)}{\arctan(\sinh(x^2)}\right)^2\to 1, \quad\left(\frac{\sinh(x^2)}{x^2}\right)^2\to 1, \quad \frac{\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{e^{\frac{-x^2}2}-\cos x } \to 1$$
we reduce to

$$\lim_{x\to 0} \frac{2\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{(\arctan(\sinh(x^2))^2}
=2\cdot \lim_{x\to 0}\frac{e^{\frac{-x^2}2}-\cos x }{x^4} = 2 \cdot \frac1{12}= \frac16$$

which can be shown by Taylor's expansion or by l'Hospital as follows
$$\lim_{x\to 0}\frac{e^{\frac{-x^2}2}-\cos x }{x^4}\stackrel{H.R.}=\lim_{x\to 0}\frac{-xe^{\frac{-x^2}2}+\sin x }{4x^3}\stackrel{H.R.}=\lim_{x\to 0}\frac{x^2e^{\frac{-x^2}2}-e^{\frac{-x^2}2}+\cos x }{12x^2}=$$
$$=\lim_{x\to 0}\frac{e^{\frac{-x^2}2}}{12}+\lim_{x\to 0}\frac{1-e^{\frac{-x^2}2}+\cos x-1 }{12x^2}=\frac1{12}+0=\frac1{12}$$
indeed by standard limits
$$\frac{1-e^{\frac{-x^2}2}+\cos x-1 }{12x^2}=\frac{1}{24}\frac{e^{\frac{-x^2}2}-1}{-\frac{x^2}2}-\frac{1}{12}\frac{1-\cos x }{x^2}=0$$
