# Limit of function with 0/0

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

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Try multiplying the top and bottom by the conjugate. – Jeel Shah Jan 20 '14 at 16:14
@cryogenic : I would start by replacing $x$ with $t^2$ to get rid of the square root. Use $\sin(t)/t \to 1$ as $t \to 0$ (which does not require either of the two things you mentioned). No more $\sin$. – Stefan Smith Jan 20 '14 at 16:18
ok so if i replace x^2 by t, I also have to add another square roots into the existing square roots? Or what you are trying to suggest is that I "divide" x^2 to have $\frac{\sin{\sqrt{x}}}{\sqrt{x}}\codt \frac{1}{x\sqrt{x}}$? – cgnx Jan 20 '14 at 16:19

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}
No, it's elementary, only asympthotic notation is required such as $\sin(x)\sim x$ and $\sqrt{1+x}\sim \frac 12x$. – Fabio Lucchini Jan 20 '14 at 16:33
Yes, it's simply a notation; for example $\sin(x)\sim x$ is another way to write $\lim_{x\to 0}\frac{\sin(x)}x=1$, while $\sqrt{1+x}-1\sim \frac 12x$ follows from $\frac{\sqrt{1+x}-1}x=\frac{1+x-1}{x(\sqrt{1+x}+1)}\to\frac 12$. – Fabio Lucchini Jan 20 '14 at 16:39