How to find this limit? $\lim\limits_{x \to 0}\frac{x-2x^3}{\sqrt{x^2-x^4}}$

Question

How to find this limit? l'Hopital's rule doesn't seem to work here...

$$\lim\limits_{x \to 0}\frac{x-2x^3}{\sqrt{x^2-x^4}}$$

Thanks!

Answer 1

Hint : $${{x-2x^3}\over\sqrt{x^2-x^4}}={{x(1-2x^2)}\over {\color{red}{|x|}\sqrt{1-x^2}}}$$. If $x>0$ you have:

$${{x-2x^3}\over\sqrt{x^2-x^4}}={{1-2x^2}\over {\sqrt{1-x^2}}}$$

and if $x<0$ you have

$${{x-2x^3}\over\sqrt{x^2-x^4}}={{2x^2-1}\over {\sqrt{1-x^2}}}$$

Answer 2

HINT: Divide up and down by $x$ and you should get non-zero denominator

Answer 3

1.Take x common & cancel it.

2.Rationalize by multiplying by the denominator's conjugate or in simple words multiply by underoot(1+x^2)/underoot(1+x^2).

3.Put value & simplfy.