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.

  • $\begingroup$ The second limit exists, just observe it is the limit of the function $2\sqrt x \cos(x)/\cos(\sqrt x)$ which approaches $0=0 \cdot 1/1,$ as $x$ approaches $0$ from the right. $\endgroup$
    – Olod
    Commented Jan 26, 2017 at 10:11
  • $\begingroup$ Hint: for small angles, the sine and the argument nearly coincide, so that the function behaves like $\sqrt x$. $\endgroup$
    – user65203
    Commented Jan 26, 2017 at 10:16

4 Answers 4


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 $$\lim_{x\to0^+}\frac{\sin x}{\sin \sqrt{x}}=\lim_{x\to0^+}\frac{\sin x}{x}\frac{\sqrt{x}}{\sin \sqrt{x}}.\sqrt{x}=1\times1\times0=0$$

  • $\begingroup$ How did you get $\frac{\sqrt{x}}{\sin \sqrt{x}} \to 1$? $\endgroup$
    – Gyakenji
    Commented Jan 26, 2017 at 10:24
  • $\begingroup$ you mentioned $\dfrac{\sin x}{x}\to0$ and this is true near zero, doesn't different from left or right $x$ tends. you can replace $\sqrt{x}$ instead of $x$. $\endgroup$
    – Nosrati
    Commented Jan 26, 2017 at 10:26
  • 1
    $\begingroup$ I don't think that $\sin(x) \approx x$ implies the first equal sign. $\endgroup$
    – MrYouMath
    Commented Jan 26, 2017 at 10:26
  • $\begingroup$ Why.? It' the same $\dfrac{\sin x}{x}$ in other words. $\endgroup$
    – Nosrati
    Commented Jan 26, 2017 at 10:27
  • $\begingroup$ Something here looks suspicious, and it is either wrong or, at least, requires proof: I can understand $\;\sin x\approx x\;$ for very small $\;x\;$ , yet then $\;\sqrt{\sin x}\approx \sqrt x\;$ ( and that only for $\;x>0\;$) which is something we do not have. Why would $\;\sin\sqrt x\approx \sqrt x\;$ ? And also that equality sign, as MrYouMath mentions, is something that'd require proof, imo $\endgroup$
    – DonAntonio
    Commented Jan 26, 2017 at 12:57

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


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 $$\lim_{x\to0^+}2\sqrt{x}$$


By a change of variable then by L'Hospital,

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


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