Limit of $\lim_{x\to0^+}\frac{\sin x}{\sin \sqrt{x}}$

Question

How do I calculate this? $$\lim_{x\to0^+}\frac{\sin x}{\sin \sqrt{x}}$$ If I tried using l'Hopital's rule, it would become $$\lim_{x\to0^+}\frac{\cos x}{\frac{1}{2\sqrt{x}}\cos \sqrt{x}}$$ which looks the same. I can't seem to find a way to proceed from here. Maybe it has something to do with $$\frac{\sin x}{x} \to 1$$ but I'm not sure what to do with it. Any advice?

Oh and I don't understand series expansions like Taylor's series.

Answer 1

By equvilency near zero $\sin x\approx x$ we have $$\lim_{x\to0^+}\frac{\sin x}{\sin \sqrt{x}}=\lim_{x\to0^+}\frac{x}{\sqrt{x}}=0$$ or $$\lim_{x\to0^+}\frac{\sin x}{\sin \sqrt{x}}=\lim_{x\to0^+}\frac{\sin x}{x}\frac{\sqrt{x}}{\sin \sqrt{x}}.\sqrt{x}=1\times1\times0=0$$

Answer 2

$$\frac{\sin x}{\sin\sqrt x}=\sqrt x\;\cdot\frac{\sin x}x\;\cdot\frac{\sqrt x}{\sin\sqrt x}\xrightarrow[x\to0^+]{}0\cdot1\cdot1=0$$

Answer 3

By a change of variable then by L'Hospital,

$$\lim_{x\to0^+}\frac{\sin x}{\sin \sqrt{x}}=\lim_{t\to0^+}\frac{\sin t^2}{\sin t}=\lim_{t\to0^+}\frac{2t\cos t^2}{\cos t}=\frac{2\cdot0\cdot1}1.$$

Answer 4

You properly wrote, after using L'Hospital once$$\lim_{x\to0^+}\frac{\cos (x)}{\frac{1}{2\sqrt{x}}\cos (\sqrt{x})}$$ which is $$\lim_{x\to0^+}2\sqrt{x}\frac{\cos (x)}{\cos( \sqrt{x})}$$ and each cosine $\to 1$. So, the limit is the same as $$\lim_{x\to0^+}2\sqrt{x}$$