Limit of function with $\frac00$

Question

I am unable to solve following limit: $$\lim_{x \rightarrow 0_+}\frac{\sin \sqrt{x}}{x^2}\left(\sqrt{x+2x^2}-\sqrt{2\sqrt{1+x}-2}\right)$$ I keep getting $ \frac{0}{0}$. I admit I haven't tried to use l'Hospital rule multiple times as the square root is not so nice to derive more than one time. Is it possible to solve this limit without using l'Hospital rule/Taylor series (which I haven't learned yet)? Thank you

Answer 1

Note that $\sqrt{2\sqrt{1+x}-2}\sim\sqrt{x+2x^2}\sim\sqrt x$, while \begin{align} (x+2x^2)-(2\sqrt{1+x}-2) &=2x^2+x+2-2\sqrt{1+x}\\ &=\frac{(2x^2+x+2)^2-4(1+x)}{2x^2+x+2+2\sqrt{1+x}}\\ &\sim\frac 94x^2 \end{align} thus \begin{align} \frac{\sin \sqrt{x}}{x^2}(\sqrt{x+2x^2}-\sqrt{2\sqrt{1+x}-2}) &=\frac{\sin \sqrt{x}}{x^2}\frac{(x+2x^2)-(2\sqrt{1+x}-2)}{\sqrt{x+2x^2}+\sqrt{2\sqrt{1+x}-2}}\\ &\sim \frac{\sqrt x}{x^2}\frac{\frac 94x^2}{2\sqrt x}\\ &=\frac 98 \end{align}