Limit calculation using Taylor series

Question

$$\lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}$$

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.

Answer 1

$$ \lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}} $$

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, $$ \lim_{x \to 0} \frac{xe^{-2x}-xe^{2x}}{\sqrt{1+x^{2}}-\sqrt{1-x^2}}=\lim_{x \to 0} \frac{x(1-2x+O(x^2))-x(1+2x+O(x^2))}{(1+\frac{1}{2}x^2+O(x^4))-(1-\frac{1}{2}x^2+O(x^4))}\to\\ \lim_{x \to 0} \frac{-4x^2+O(x^3)}{x^2+O(x^4)}=\lim_{x \to 0} \frac{-4+O(x)}{1+O(x^2)}=-4 $$

Answer 2

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

$xe^{-2x}-xe^{2x}=-4x^2+x^2\epsilon(x)$

Answer 3

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 $\sqrt{1\pm x^2}=1+O(x^2)$.

Answer 4

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.