# Find the limit analytically when the sine functions have square roots?

Find the limit analytically of the following:

$\lim \limits_{x \to 0} \frac {\sin(\sqrt{2x})}{\sin(\sqrt{5x})}$

The closes thing we learned in class about this was that $\sin(x)$ over $x$ will equal $1$, but I'm not sure how I can apply this here. Note that Wolfram Alpha tells me to use L'Hopital's Rule, but we haven't even learned that or even heard of that in class yet, and this is due Monday, so I doubt we'll learn in the coming day.

Is there any way I can solve this without L'Hopital's Rule or do I just go ahead and teach it to myself? Note that we are in the first chapter of Calculus so please don't bring complicated methods that a first time Calculus student wouldn't know.

For $a$ and $b$ positive numbers we have \begin{align*} \lim_{x\to 0^+}\frac{\sin \sqrt{ax}}{\sin \sqrt{bx}}=\lim_{x\to 0^+}\frac{\frac{\sqrt{a}\sin \sqrt{ax}}{\sqrt{ax}}}{\frac{\sqrt{b}\sin \sqrt{bx}}{\sqrt{bx}}} &=\frac{\sqrt{a}}{\sqrt{b}}\lim_{x\to 0^+}\frac{\frac{\sin \sqrt{ax}}{\sqrt{ax}}}{\frac{\sin \sqrt{bx}}{\sqrt{bx}}}=\frac{\sqrt{a}}{\sqrt{b}}\frac{\lim_{x\to 0^+}\frac{\sin \sqrt{ax}}{\sqrt{ax}}}{\lim_{x\to 0^+}\frac{\sin \sqrt{bx}}{\sqrt{bx}}}=\frac{\sqrt{a}}{\sqrt{b}}\cdot\frac{1}{1}=\sqrt{\frac{a}{b}} \end{align*}
• @JozemiteApps you can always multiply the numerator and denominator by any factor, no matter what that factor is, as long as you take care to exclude cases where the factor is zero. That's not a concern here because $x > 0$. – David Z Aug 31 '15 at 7:49
What the limit $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$ really means is that $\sin x$ behaves like $x$ around $0$. Therefore, as we approach $0$, $$\frac{\sin(\sqrt{2x})}{\sin(\sqrt{5x})} \sim \frac{\sqrt{2x}}{\sqrt{5x}}=\sqrt{\frac{2}{5}}.$$