Evaluate $\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)}$ without L'Hopital

Question

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!

Answer 1

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

Answer 2

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:

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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