## Bomull Group Title lim(x->0) sin(x)/sin(2x) ? one year ago one year ago

1. Bomull

$\lim (x->0) \frac{ \sin(x) }{ \sin(2x) }$ ?

2. ash2326

$\lim_{x\to 0}\frac{\sin x }{\sin 2x}$ We know that $\lim_{x\to 0}\frac{\sin x }{x}=1$ Could you use this here to solve ? @Bomull

3. Bomull

I don't know how..

4. ash2326

Okay, I'll explain you $\lim_{x\to 0} \frac{\sin x}{\sin 2x}$ Let's multiply and divide by x $\lim_{x\to 0} \frac{\sin x}{\sin 2x}\times \frac x x$ $\lim_{x\to 0} \frac{\sin x}{x}\times \frac{x}{\sin 2x}$ Do you understand till here?

5. Bomull

yeah

6. ash2326

$\lim_{x\to 0} \frac{\sin x}{x}\times \frac{x}{\sin 2x}$ Let's multiply and divide by 2 $\lim_{x\to 0} \frac{\sin x}{x}\times \frac{2x}{\sin 2x}\times \frac 1 2$ We know that $\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{nx}{\sin nx}=1$ so we get $\lim_{x\to 0} \frac{\sin x}{x}\times \frac{2x}{\sin 2x}\times \frac 1 2$ $1\times 1 \times \frac 1 2=\frac 1 2$

7. ash2326

Do you understand this?

8. Bomull

ah ok thanks! $\lim_{x \rightarrow 0} \frac{ nx }{ \sin nx }$ is good to know. I think I got it

9. ash2326

Good :)