Question

lim(x->0) sin(x)/sin(2x) ?

Answer 1

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

Answer 2

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

Answer 3

I don't know how..

Answer 4

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?

Answer 5

yeah

Answer 6

\[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{x}{\sin 2x}\]
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
\[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{2x}{\sin 2x}\times \frac 1 2\]
\[1\times 1 \times \frac 1 2=\frac 1 2\]

Answer 7

Do you understand this?

Answer 8

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

Answer 9

Good :)