Question

Bantuin dong ( ngerjain nya pake cara ya klo bisa si doto biar saya jelas )

mengenai limit trigono dan aljabar

Answer 1

1) lim(x->0) (1- cos 2x)(cos 2x)/ (tan 5x. sin 2x)
lim (x->0)  (2 sin x sin x)(cos 2x ) /tan 5x sin 2x)
= 2(x)(x)/(5x)(2x)  . cos 2x
= 1/5 . cos 2(0)
= 1/5 (1)
= 1/5

2) lim(x->0)  tan 4x/ tan 8x = 4/8 = 1/2

3) lim(x->1) (x+2) . sin (x-1)/ (x^2 + 2x -3) =
= sin(x-1)/(x-1) . (x+2)/(x+3)
= (x+2)/(x+3)
x = 1 --> (1+2)/(1+3) = 3/4