solving limits

Question

lim      sin2x

x->0       x

Answer 1

lim_x→0(sinx/x) = 1

 

So, lim_x→0[sin(2x)/x] = 2 lim_x→0[sin(2x) / (2x)]

 

                                            Let θ = 2x.  As x→0, θ→0.

 

                               = 2lm_θ→0[sinθ/θ] = 2(1) = 2

 

 

Answer 2

I assume you meant to say

 

lim_x→0 [sin(2x)/x] = ?

 

Use L'Hopital's rule.

 

lim_x→0 [sin(2x)/x] = lim_x→0 [2cos(2x)/1] = 2cos(0) = 2