3
$\begingroup$

Here is my approach: $$\lim _{x\to 0^+}\frac{\sin 3x}{\sqrt{1-\cos ^3x}}=\lim _{x\to 0^+}\frac{1}{\sqrt{\cos^2x+\cos x+1}}\times\frac{\sin 3x}{\sqrt{1-\cos x}}$$ $$=\lim _{x\to 0^+}\frac1{\sqrt3}\times\frac{\sin 3x}{\sqrt{1-\cos x}}\times\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}=\frac{\sqrt2}{\sqrt3}\times\lim _{x\to 0^+}\frac{\sin 3x}{\sin x}=\sqrt6$$

Can you evaluate $\lim _{x\to 0^+}\dfrac{\sin 3x}{\sqrt{1-\cos ^3x}}$ with other approaches?

$\endgroup$
3
$\begingroup$

You can apply L'Hopital's Rule:$$\lim_{x\to0^+}\frac{\sin^2(3x)}{1-\cos^3(x)}=\lim_{x\to0^+}\frac{6\sin(3x)\cos(3x)}{3\sin(x)\cos^2(x)}.\tag1$$And now you can prove (again by L'Hopital's Rule) that $\lim_{x\to0^+}\frac{\sin(3x)}{\sin(x)}=3$, and it will now follow that $(1)$ is equal to $6$.


Edit: Actually, you don't have to use L'Hopital's Rule for a second time in order to compute $\lim_{x\to0^+}\frac{\sin(3x)}{\sin(x)}=3$. It follows from the fact that\begin{align}\sin(3x)&=\sin(x+2x)\\&=\sin(x)\cos(2x)+\cos(x)\sin(2x)\\&=\sin(x)\cos(2x)+2\sin(x)\cos^2(x),\end{align}and that therefore$$\lim_{x\to0^+}\frac{\sin(3x)}{\sin(x)}=\lim_{x\to0^+}\bigl(\cos(2x)+2\cos^2(x)\bigr)=3.$$

$\endgroup$
2
  • $\begingroup$ Thank you very much! I've already got two great answers and I don't know which one should I mark as accepted one. $\endgroup$ Jun 21 at 20:46
  • $\begingroup$ The one which closer to your taste. $\endgroup$ Jun 21 at 21:03
6
$\begingroup$

I would write something like that,

$$\sin(3x) \sim 3x$$

$$\cos^3(x) = \left(1-\dfrac{x^2}{2}+o(x^2)\right)^3 = 1-3\dfrac{x^2}{2}+o(x^2)$$

Hence,

$$1-\cos^3(x) \sim 3\dfrac{x^2}{2} $$

Hence,

$$\sqrt{1-\cos^3(x)} \sim \sqrt{3/2}x $$

Finally,

$$\dfrac{\sin(3x)}{\sqrt{1-\cos^3(x)}} \sim \dfrac{3}{\sqrt{3/2}} \sim \sqrt{6}$$

$\endgroup$
1
  • 1
    $\begingroup$ This is the approach I'd use, though it's worth being explicitly careful about the error terms. Taylor series are by far the easiest way for my brain to think about these sorts of problems. $\endgroup$ Jun 21 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.