Question

I have tried doing this several different ways but can't get the answer: lim x->0 of -4x/sin(2x) + x/cos(2x) Thanks!

Answer 1

can you do this part ? x/cos(2x)

Answer 2

write the first fraction as
\[ -2 \frac{1}{\frac{\sin(2x)}{2x}} \]

Answer 3

You do know that the limit of a sum is the sum of the individual limits ?
lim x->0 -4x/sin(2x) + lim x->0 x/cos(2x)

when x is 0, cos(2x)= cos(0)= 1
and 0/1 = 0
so lim x->0 x/cos(2x) = 0

now do the left fraction. btw, the limit u->0 sin(u)/u = 1