1
$\begingroup$

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.

  • 0
    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. – 2017-01-26
  • 0
    Hint: for small angles, the sine and the argument nearly coincide, so that the function behaves like $\sqrt x$. – 2017-01-26

4 Answers 4

7

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

  • 0
    How did you get $\frac{\sqrt{x}}{\sin \sqrt{x}} \to 1$? – 2017-01-26
  • 0
    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 $x$. – 2017-01-26
  • 1
    I don't think that $\sin(x) \approx x$ implies the first equal sign. – 2017-01-26
  • 0
    Why.? It' the same $\dfrac{\sin x}{x}$ in other words. – 2017-01-26
  • 0
    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 – 2017-01-26
  • 0
    @DonAntonio Is this really a simple substitution or merely is a intuition which force us to do this. For a beginner - who we solve a problem for him - this sufficient to say $\sin x\approx x$ as $x\to0$, but the main idea comes from here $$\lim_{x\to0}\dfrac{\sin x}{x}=\dfrac{\lim_{x\to0}\sin x}{\lim_{x\to0}x}=1$$ because the limits in numerator and denominator exist. This lead us $\displaystyle\lim_{x\to0}\sin x=\lim_{x\to0}x$ thus our purpose simply interpreted as which we say $\sin x\approx x$ as $x\to0$. – 2017-01-26
4

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

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

3

By a change of variable then by L'Hospital,

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