Limit calculation using Taylor series $$\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}$$
Any tip about denominator, for numerator i am trying to plug in the value $-2x$ and $2x$ in pre-computed series of $e^x$. I have to calculate the limit using taylor series.
 A: $$
\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}
$$
Use Taylor expansions:
$$
e^{-2x}=1-2x+O(x^2)\\
e^{2x}=1+2x+O(x^2)\\
\sqrt{1+x^2}=1+\frac{1}{2}x^2+O(x^4)\\
\sqrt{1-x^2}=1-\frac{1}{2}x^2+O(x^4)
$$
Then, 
$$
\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}=\lim_{x \to 0} \frac{x(1-2x+O(x^2))-x(1+2x+O(x^2))}{(1+\frac{1}{2}x^2+O(x^4))-(1-\frac{1}{2}x^2+O(x^4))}\to\\
\lim_{x \to 0} \frac{-4x^2+O(x^3)}{x^2+O(x^4)}=\lim_{x \to 0}  \frac{-4+O(x)}{1+O(x^2)}=-4
$$
A: Hint for numerator:
We have
$e^X=1+X+X\epsilon_0(X)$
thus
$e^{2x}=1+2x+x\epsilon_1(x)$
and
$e^{-2x}=1-2x+x\epsilon_2(x)$
which give
$xe^{-2x}-xe^{2x}=-4x^2+x^2\epsilon(x)$
A: HINT:
Note that we can write
$$\frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=-2\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)\left(\frac{\sinh(2x)}{2x}\right)$$
and the Taylor series of $\sinh(2x)=2x+O(x^3)$ while the Taylor series of $\sqrt{1\pm x^2}=1+O(x^2)$.
A: As $u\to 0, e^u = 1 +u +o(u), (1+u)^{1/2} = 1 + u/2 +o(u).$ You can consider these equalities a case Taylor if you like, but it's really just the definition of the derivative at $0$ for these functions.
Thus $xe^{-2x} = x(1 -2x +o(-2x)) = x-2x^2+o(x^2),$ and the same kind of thing for each term in the given expression. The limit of $-4$ will fall right out.
