<s>[INST] 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.
 [/INST] $$
\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}
$$
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, 
$$
\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}=\lim_{x \to 0} \frac{x(1-2x+O(x^2))-x(1+2x+O(x^2))}{(1+\frac{1}{2}x^2+O(x^4))-(1-\frac{1}{2}x^2+O(x^4))}\to\\
\lim_{x \to 0} \frac{-4x^2+O(x^3)}{x^2+O(x^4)}=\lim_{x \to 0}  \frac{-4+O(x)}{1+O(x^2)}=-4
$$
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