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Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$?

The justification is that $\lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$

But, I am not seeing the connection.
L'Hospital's rule? Is there a double angle substitution happening?

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    $\begingroup$ I would like to point out that the use of L'Hopital's rule to evaluate $\lim_{x\to 0} \frac{\sin(x)}{x}$ is circular, since it requires the knowledge of the derivative of $\sin(x)$ at zero, which is what $\lim_{x\to0} \frac{\sin(x)}{x}$ is in the first place. Therefore, either accept and use the fact that $\lim_{x\to 0} \sin(x)/x = 1$ or prove it in some other fashion. There are many geometric and analytic proofs of the limit out there, but likely it is expected for you to memorize the limit. $\endgroup$ – Joel Jan 14 '16 at 14:57

10 Answers 10

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$$\frac{\sin(6x)}{{\sin(2x)}} =\frac{\sin(6x)}{6x} \cdot \frac{2x}{{\sin(2x)} }\cdot \frac{6}{2} $$

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You don't need to know any special limits or derivatives, you can do it with trig identities:

From

$$\sin3\theta=\sin(\theta+2\theta)=\sin\theta\cos2\theta+\cos\theta\sin2\theta=\sin\theta\cos2\theta+2\sin\theta\cos^2\theta$$

we have, letting $\theta=2x$,

$${\sin6x\over\sin2x}={\sin2x(\cos4x+2\cos^22x)\over\sin2x}=\cos4x+2\cos^22x$$

and thus

$$\lim_{x\to0}{\sin6x\over\sin2x}=\lim_{x\to0}(\cos4x+2\cos^22x)=\cos0+2\cos^20=1+2=3$$

(The final evaluation relies, of course, on knowing that the cosine function is continuous.)

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  • $\begingroup$ I hadn't thought of doing it like this. Innovative ! $\endgroup$ – user230452 Jan 21 '16 at 10:44
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As user Daniel has written you can express the ratio $\frac{\sin 6x}{\sin 2x}$ in a way which makes it amenable to the use of standard limit $$\lim_{x\to 0} \frac{\sin x}{x} = 1\ .$$ Notice that $$\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\sin 2x}{2x} = \lim_{x\to 0} \frac{\sin 6x}{6x}\ .$$ As long as the argument $x$ is not equal to zero (and $\sin 2x \neq 0$) you can prolong the ratio $\frac{\sin 6x}{\sin 2x}$ as follows.

$$\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{6x}{2x} \cdot \frac{2x}{\sin 2x} = \underbrace{\frac{\sin 6x}{6x}}_{\to 1} \cdot \frac{6}{2} \cdot \underbrace{\frac{1}{\frac{\sin 2x}{2x}}}_{\to\frac{1}{1}} \to 1 \cdot 3 \cdot 1 = 3 \ .$$ Hence you obtain the result $$\lim_{x\to 0}\frac{\sin 6x}{\sin 2x} = 3 \ .$$

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$$\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)}$$ $$= \lim_\limits{x\to 0}\frac{6}{2} \cdot \frac{\frac{\sin(6x)}{6x}}{\frac{\sin(2x)}{2x}}$$ $$=\frac{6}{2}$$ $$=3$$

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HINT: rewrite the quotient in the form $$\frac{\sin(6x)}{6x}\frac{1}{\frac{\sin(2x)}{2x}}\cdot 3$$

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Since the limit of both the numerator and denominator are 0 use L'Hopitals rule. You then get $\lim_{x \to 0} \frac{6cos(6x)}{2cos(2x)}=\frac{6}{2}=3$

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  • $\begingroup$ "Since the limit of both the numerator and denominator are 0 use L'Hopitals rule" No thanks, see comment on main. $\endgroup$ – Did Jan 14 '16 at 15:42
  • $\begingroup$ Good call. My bad. $\endgroup$ – siegehalver Jan 14 '16 at 15:45
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The key is to use Hopital rule: $$\begin{align} \lim_\limits{x\to 0}\frac{\sin(x)}{x} &= \lim_\limits{x\to 0}\frac{\sin'(x)}{1}\\ &= \lim_\limits{x\to 0}\frac{\cos(x)}{1}\\ &= \lim_\limits{x\to 0}\frac{1}{1}\\ &=1 \end{align} $$ hence $$ \begin{align} \lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} &= \lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} \cdot \frac{2x}{6x} \cdot \frac{6}{2}\\ &= \lim_{x\to 0} \frac{\sin(6x)}{6x} \cdot \frac{\sin(2x)}{6x} \cdot 3\\ &= 1 \cdot 1 \cdot 3\\ &=3 \end{align}$$

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  • $\begingroup$ "The key is to use Hopital rule" Well, no, see comment on main. $\endgroup$ – Did Jan 14 '16 at 15:41
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$$\lim_{x\to0}\frac{2x}{\sin2x}\cdot\frac{\sin6x}{6x}=1$$

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In asymptotic terms, $\displaystyle \lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$ means that "$\sin(x)$ behaves like $x$ around $0$".

To this understanding, $\sin(6x)$ behaves like $6x$ and $\sin(2x)$ behaves like $2x$ around $0$.

Therefore, $\displaystyle \frac{\sin(6x)}{\sin(2x)}$ behaves like $\displaystyle \frac{6x}{2x}=3$ around $0$.

In other words, $\displaystyle \lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = 3.$

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Taking a look at this page we find

$$1+2\cos u + 2 \cos 2 u + 2 \cos 3 u + ... + 2 \cos n u=\frac{\sin \frac{2n+1}{2}u}{\sin\frac u2}$$

In the left hand side set $u=4x$ and $n=1$ so that the left hand side becomes $$\frac{\sin 6x}{\sin 2x},$$ therefore we have $$1+2\cos 4x = \frac{\sin 6x}{\sin 2x}$$

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