Question

Tolong dijawab dg caranya!!thanks
1. lim (x tan 2x/1-cos 6x)
x-->0

2. Jika f(x)=√x+1 dan (f o g)(x)=2√x-1, maka g(x)=?
a. 4x-5
b. 4x-3
c. 5x-4
d. 2x-3
e. 2x-1

3. lim (x³+3x²+2x/(x²-1)sin3x)
x--> 0

4. lim (3x²-10x+3)/cos(x-3) tg(x-3)
x -->3

Answer 1

1) lim (x->0)  xtan 2x / (1 - cos6x)
(x tan 2x ) / (2 sin² 3x) = (x)(2x) / (2)(3x)(3x) = 1/9

2) f{ g(x)} = fog(x)
√(g(x) +1) = 2√(x - 1)
√(g(x) +1) = √4(x - 1)
g(x)+1 = 4x - 4
g(X) = 4x - 5

3)
lim(x->0) ( x³+3x² + 2x)/(x²-1)(sin 3x)
lim x->0  x (x² + 3x +2) / (x-1)(x+1)(sin 3x)
lim(x->0) x(x+1)(x+2) / (x-1)(x+1)(sin 3x)
lim(x->0) x (x+2) / (x-1). sin 3x
lim(x->0) x / sin 3x . lim (x->0) (x+2)/(x-1)
= 1/3 . (2/-1)
= 1/3 (-2)
 = - 2/3

4) lim(x->3) { 3x² -10x +3)/ cos(x-3. tg (x-3)
lim(x->3) (3x -1)(x - 3) / cos(x-3),tg(x-3)
lim(x->3) (x-3)/tg(x-3) . lim(x->3) (3x-1)/cos(x-3)
= 1 . (8)/cos0
= 8/1
= 8