Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Thanks!

share|improve this question
    
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
1  
$${\sin6x\over x}\div{\sin2x\over x}$$ –  Gerry Myerson Nov 13 '13 at 11:54

7 Answers 7

up vote 10 down vote accepted

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

share|improve this answer

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.

HINT:

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

share|improve this answer

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

share|improve this answer
1  
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

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 $$\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)} = \frac{6}{2} = 3 $$

share|improve this answer

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

share|improve this answer

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

share|improve this answer

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 $\lim\limits_{t \to 0}\,\dfrac{\sin t}{t} = 1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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