# Limits question math help?

Question : Find $$\lim_{x\to\pi}\frac{\sin(3x)}{\sin(2x)}$$ If I divide by $x$ in the denominator and the numerator, I still get no result.

Should I replace $x-\pi=0$?

By the way,I shouldnt use L'hopital...

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Yes...that would work. –  Tapu Feb 4 '13 at 13:38
No L'Hospital... Are you allowed to use $\lim_{u\rightarrow 0}\sin (u)/u=1$? –  1015 Feb 4 '13 at 14:05
the limit $\displaystyle \lim_{t\to 0} \frac{\sin t}{t} =1$ can be proven without LH : see this –  aziiri Feb 4 '13 at 14:14
@aziiri Yes. That is why I think the OP should tell us if he/she's allowed to use this limit, or if a proof of this is required too. –  1015 Feb 4 '13 at 14:22
Sounds like homework by the way; should be tagged as such. –  Noldorin Feb 4 '13 at 17:52

By angle sum formula for sine, we have $$\sin(3x)=\sin(2x)\cos(x)+\sin(x)\cos(2x),$$ and double-angle formula for sine tells us that $$\sin(2x)=2\sin(x)\cos(x),$$ so we have \begin{align}\frac{\sin(3x)}{\sin(2x)} &= \frac{\sin(2x)\cos(x)+\sin(x)\cos(2x)}{\sin(2x)}\\ &= \frac{2\sin(x)\cos^2(x)+\sin(x)\cos(2x)}{2\sin(x)\cos(x)}\\ &= \frac{\left(2\cos^2(x)+\cos(2x)\right)\sin(x)}{2\sin(x)\cos(x)}\\ &= \frac{2\cos^2(x)+\cos(2x)}{2\cos(x)},\end{align} so long as $\sin(x)\neq 0$. For $x$ sufficiently close to (but not equal to) $\pi$, we have that $\sin(x)\neq 0$. Thus, $$\lim_{x\to\pi}\frac{\sin(3x)}{\sin(2x)}=\lim_{x\to\pi}\frac{2\cos^2(x)+\cos(2x)}{2\cos(x)}=\frac{2\cos^2(\pi)+\cos(2\pi)}{2\cos(\pi)}=\frac{2(-1)^2+(1)}{2(-1)}=-\frac32.$$

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Hint: Let $x=\pi +h$ and note that $$\lim_{x\to 0} \frac{\sin(x)}{x}=1$$

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L'Hôpital's rule is your friend here. This is by far the simplest approach; no substitution or trigonometric expansions are required. That should be enough of a hint.

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No substitution needed, L'Hospital works at $\pi$ too. –  1015 Feb 4 '13 at 14:03
@julien: You're right, I'm not sure why I suggested that. Thanks for pointing it out. ;-) –  Noldorin Feb 4 '13 at 17:53

$$\lim_{x\to\pi}\frac{\sin 3x}{\sin 2x}=\lim_{y\to0}\frac{\sin(3\pi+3y)}{\sin(2\pi+2y)}=\lim_{y\to0}\frac{-\sin(3y)}{\sin(2y)}=-\lim_{y\to0}\frac{\sin(3y)}{3y}\frac{2y}{\sin(2y)}.\frac{3}{2}=-\frac{3}{2}$$

If you know L. Hospital's rule, then it is very easy.

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Sorry, I did not see your restriction on LH rule. –  Tapu Feb 4 '13 at 13:47

$$\frac{\sin3x}{\sin2x}=\frac{3\sin x-4\sin^3x}{2\sin x\cos x}=\frac{3-4\sin^2x}{2\cos x} \text{ if } \sin x\ne0$$

As $x\to\pi,\sin x\to\sin \pi\implies \sin x\to 0\implies \sin x\ne0$

So, $$\lim_{x\to \pi}\frac{\sin3x}{\sin2x}=\lim_{x\to \pi}\frac{3\sin x-4\sin^3x}{2\sin x\cos x}=\lim_{x\to \pi}\frac{3-4\sin^2x}{2\cos x}=-\frac32 \text{ as }\cos\pi=-1,\sin\pi=0$$

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This was originally posted (and then deleted) by another user. I fine-tuned it slightly (since it was a decent answer) and posted as a community wiki answer (since it certainly wasn't mine).

We have $\sin (3x)=-\sin(3x-3\pi)$ and $\sin(2x)=\sin (2x-2\pi)$. So $$\frac{\sin(3x)}{\sin(2x)}=-\frac{\sin(3(x-\pi))}{\sin(2(x-\pi))}=-\frac{\sin (3u)}{\sin (2u)}$$ where $u=x-\pi$ now tends to $0$.

Now use the famous $$\lim_{y\rightarrow 0}\frac{\sin y}{y}=1$$ and the fact that $$-\frac{\sin (3u)}{\sin (2u)}=-\frac{3}{2}\frac{\sin (3u)}{3u}\frac{2u}{\sin(2u)}$$ to get your limit of $-3/2$.

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