Limit trigonometrimohon bantuannya terimakasihh

Limit trigonometrimohon bantuannya terimakasihh

Jawab:

Penjelasan dengan langkah-langkah:

limit trigonometri

lim x->0 sin x / x= 1

soal

$sf lim_{xto 0}~ dfrac{sin^3 x}{sqrt{4x+16} - 4}$

$sf= lim_{xto 0}~ dfrac{sin3 x~(sqrt{4x+16} + 4)}{4x}$

$sf= lim_{xto 0}~ dfrac{sin3 x~(4+ 4)}{4x}$

$sf= lim_{xto 0}~ dfrac{3(4+ 4)}{4} = 6$

soal k.

$sf= lim_{xto 0}~ dfrac{x { cos^2 6x- 1}}{sin~ 3x.~ tan^2 ~2x}$

$sf= lim_{xto 0}~ dfrac{x { -2 sin^2 3x}}{sin~ 3x.~ tan^2 ~2x}$

$sf= lim_{xto 0}~ dfrac{-2x (3x)^2}{3x (2x)^2} = dfrac{-2(9)}{3(4)} = -dfrac{3}{2}$

soal l

$sf lim_{xto 0}~ dfrac{2- 2cos ~x + cos~ 2x}{x^2}$

$sf = lim_{xto 0}~ dfrac{2- 2cos ~x + 1- 2sin^2 x}{x^2}$

$sf = lim_{xto 0} ~(2- 2cos ~x + 1) - lim_{xto 0}~ dfrac{ 2sin^2 x}{x^2}$

$sf = {2- 2(1)+ 1} - dfrac{ 2x^2}{x^2}$

$sf = (1) - (2) = - 1$