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I am trying to evaluate the following limit without L'Hopital's:

$$\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)}$$

I know I have to use the fact that $\frac{\sin x}{x} = 1$ but I don't know how to get the limit from the above to $\frac{\sin x}{x}$ or even a portion of it to that. I know how to evaluate limits like the following

$$\lim_{x \rightarrow 0}\frac{\sin 3x}{x} = 3$$

if that's any help. Any hints would be appreciated


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When $x\sim 0$ we know that $\sin(ax)\sim ax$. – Babak S. Nov 13 '13 at 11:53
So, I guess the answer is just $\frac{6x}{2x} = 3$? – Jeel Shah Nov 13 '13 at 11:53
$${\sin6x\over x}\div{\sin2x\over x}$$ – Gerry Myerson Nov 13 '13 at 11:54
up vote 14 down vote accepted

hint : $$\frac{\sin6x}{\sin 2x}= \frac{\frac{\sin6x}{6x}}{\frac{\sin 2x}{2x}}\cdot 3 $$

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It’s not true that $\dfrac{\sin x}x=1$; you mean the fact that $\lim\limits_{x\to 0}\dfrac{\sin x}x=1$. The distinction is important.


$$\frac{\sin 6x}{\sin 2x}=\frac{\sin 6x}{6x}\cdot\frac{3\cdot2x}{\sin 2x}$$

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Since $\sin x\sim x$ as $x\to0$ , it follows that $\lim_{x\to0}\frac{\sin6x}{\sin2x}=\lim_{x\to0}\frac{6x}{2x}=\frac62=3$.

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I don't think it is entirely correct to say that $\sin x \to x$ as $x \to 0$. You should probably say $\sin x \sim x$ as $x \to 0$. – Dan Shved Nov 13 '13 at 12:26
@DanShved: Not entirely correct? It is totally meaningless lol! – Taladris Nov 13 '13 at 13:15

This is a different take on this easy problem. Let $t = 2x$ so that $t \to 0$ as $x \to 0$. Then we can see that $$\begin{aligned}L &=\lim_{x \to 0}\frac{\sin 6x}{\sin 2x}\\ &= \lim_{t \to 0}\frac{\sin 3t}{\sin t}\\ &= \lim_{t \to 0}\frac{3\sin t - 4\sin^{3}t}{\sin t}\\ &= \lim_{t \to 0}(3 - 4\sin^{2}t) = 3\end{aligned}$$ Note that the above uses a much simpler result $\lim\limits_{t \to 0}\,\sin t = 0$ compared to the slightly harder result $\lim\limits_{t \to 0}\,\dfrac{\sin t}{t} = 1$.

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We can write $$\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)}$$ as follows:

$$\lim_{x\rightarrow 0} \frac{\sin(6x)\frac{6}{6}}{\sin(2x) \frac{2}{2}{}} $$

We already know that $\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1 $

The same applies for $\lim_{x\rightarrow 0} \frac{\sin(6x)}{6x}$ and

$\lim_{x\rightarrow 0} \frac{\sin(2x)}{2x}$ because, if you set a new variable $y = 6x$ and $2x$ respectively, since if $x\rightarrow 0$ then $2x\rightarrow 0$ and $6x\rightarrow 0$ as well.

So from there we get that $$\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)} = \frac{6}{2} = 3 $$

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$$\frac{\sin6x}{\sin2x}=3\frac{\sin 6x}{6x}\frac{2x}{\sin 2x}\xrightarrow[x\to 0]{}3\cdot 1\cdot 1=3$$

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The most important point of $\displaystyle\lim_{h\to0}\frac{\sin (h)}h=1$ is :

the limit variable, the angle of sine (in radian) and the denominator must be same.

We can use the Product Rule of Limits to ensure the above constraint

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