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
 A: If you are aware of Taylor expansions then:
$$
e^{\frac{-x^2}{2}} =_{x \rightarrow 0} 1-\frac{x^2}{2} + \frac{x^4}{8} + o(x^4)
$$
$$
\cos(x) =_{x \rightarrow 0} 1 - \frac{x^2}{2} + \frac{x^4}{4!} + o(x^4)
$$
Then
$$
e^{\frac{-x^2}{2}} - \cos(x) =_{x \rightarrow 0} \frac{x^4}{12} + o(x^4)
$$
Now because $sin(x) =_{x \rightarrow 0} x + o(x)$ then
$$ 
2 \sin( e^{\frac{-x^2}{2}} - \cos(x)) =_{x \rightarrow 0} \frac{x^4}{6} + o(x^4)
$$
And for the denominator:
$$
\sinh(x^2) =_{x \rightarrow 0} x^2 + o(x^2)
$$
$$
\arctan(x) =_{x \rightarrow 0} x + o(x)
$$
So:
$$
(\arctan(\sinh(x^2))^2 =_{x \rightarrow 0} x^4 + o(x^4)
$$
And thus:
$$
\frac{2 \sin( e^{\frac{-x^2}{2}} - \cos(x))}{(\arctan(\sinh(x^2))^2} =_{x \rightarrow 0} \frac{1}{6} + o(1)
$$
A: 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
$$\frac{1-e^{\frac{-x^2}2}+\cos x-1 }{12x^2}=\frac{1}{24}\frac{e^{\frac{-x^2}2}-1}{-\frac{x^2}2}-\frac{1}{12}\frac{1-\cos x }{x^2}=0$$
A: It might be easier to consider some common limits like
$$
\lim_{x\to0}\frac{\sin(x)}{x}=1\tag1
$$
$$
\lim_{x\to0}\frac{\sinh(x)}{x}=1\tag2
$$
$$
\lim_{x\to0}\frac{\arctan(x)}{x}=1\tag3
$$
Using these three limits, we get
$$
\begin{align}
&\lim_{x\to0}\frac{2\sin\left(e^{-x^2/2}-\cos(x)\right)}{\left(\arctan\left(\sinh\left(x^2\right)\right)\right)^2}\\
&=\scriptsize2\,\underbrace{\lim_{x\to0}\frac{\sin\left(\color{#C00}{e^{-x^2/2}-\cos(x)}\right)}{\color{#C00}{e^{-x^2/2}-\cos(x)}}\vphantom{\frac1{\left(x^2\right)}}}_{1}\underbrace{\left(\lim_{x\to0}\frac{\color{#090}{\sinh\left(x^2\right)}}{\arctan\left(\color{#090}{\sinh\left(x^2\right)}\right)}\lim_{x\to0}\frac{\color{#00F}{x^2}}{\sinh\left(\color{#00F}{x^2}\right)}\right)^2}_{(1\cdot1)^2}\lim_{x\to0}\frac{e^{-x^2/2}-\cos(x)}{x^4}\tag4\\
&=2\lim_{x\to0}\frac{e^{-x^2/2}-\cos(x)}{x^4}\tag5
\end{align}
$$
which is a lot easier to evaluate with L'Hôpital:
$$
\begin{align}
2\lim_{x\to0}\frac{e^{-x^2/2}-\cos(x)}{x^4}
&=2\lim_{x\to0}\frac{-xe^{-x^2/2}+\sin(x)}{4x^3}\tag6\\
&=2\lim_{x\to0}\frac{\left(x^2-1\right)e^{-x^2/2}+\cos(x)}{12x^2}\tag7\\
&=2\lim_{x\to0}\frac{\left(-x^3+3x\right)e^{-x^2/2}-\sin(x)}{24x}\tag8\\
&=2\lim_{x\to0}\frac{\left(x^4-6x^2+3\right)e^{-x^2/2}-\cos(x)}{24}\tag9\\[3pt]
&=\frac16\tag{10}
\end{align}
$$
