Question

find the limit as x approaches 0 of sin2x/4x

Answer 1

\[\lim_{x\to0} \frac{\sin{2x}}{4x}=\lim_{x\to0} \frac{\sin{x}\cos{x}}{2x}=\lim_{x\to0}\frac{\sin{x}}x\times\lim_{x\to0}\frac{cos x}{2}=1\times\frac12=\frac12\]

Answer 2

I don't undedrstand how you got sinx cosx?

Answer 3

It's a trigonometric identity; as you know...\[\sin{\alpha+\beta}=\sin\alpha\cos\alpha+\cos\alpha\sin\beta\]Now, since we have...\[\sin{2x} = \sin{x+x}\]... we can apply that earlier identity to derive as follows.\[\sin{2x}=\sin{x+x}=\sin{x}\cos{x}+\cos{x}\sin{x}=2\sin{x}\cos{x}\]

Answer 4

Ooops, I meant...\[\sin{\alpha+\beta}=\sin\alpha\cos\beta+\cos\alpha\sin\beta\]

Answer 5

i think u may change the numerator : 4x=2x*2
so, lim (x->0) sin2x/4x
= lim (x->0) sin2x/(2x*2)
= lim (x->0) (sin2x/2x)*(1/2)
and we knowed that :
lim (x->0) sin2x/2x = 1

Answer 6

and finally answer is 1 * 1/2 = 1/2