Find the limit of x as it approaches zero. (Sin^2 8x)/x^2

Answer 1

lim ((sin 8x)^2)/x^2 as x->0
since as x->0 the denominator becomes zero, use l'hopital's rule:
lim (16 sin(8x)cos(8x) / 2x)
lim (8 sin(16x))/2x
another l'hopital's rule:
lim (128 cos (16x))/2
as x->0,
(128 * cos(0))/2
= 64

Answer 2

as x --> 0 sin (8x)^2---> 64 x^2
64 x^2/x^2 = 64