kak tolong bantu jawab dikumpulkan nanti sore
LImit bentuk 0/0
TRigonometri
Penjelasan dengan langkah-langkah:
[tex]\sf 15. lim_{x\to 0}\ \dfrac{\sin x -\tan x}{x^2. \tan x}[/tex]
(*kalikan cos x/cos x )
[tex]\sf = lim_{x\to 0}\ \dfrac{\sin x -\tan x}{x^2. \tan x} . \times \dfrac{\cos x}{\cos x}[/tex]
[tex]\sf = lim_{x\to 0}\ \dfrac{\sin x \cos x - \sin x }{x^2. \sin x} .[/tex]
[tex]\sf = lim_{x\to 0}\ \dfrac{\sin x (\cos x -1)}{x^2. \sin x} .[/tex]
[tex]\sf = lim_{x\to 0}\ \dfrac{\sin x (1- 2\sin^2 \frac{1}{2}x - 1)}{x^2. \sin x} .[/tex]
[tex]\sf = lim_{x\to 0}\ \dfrac{\sin x (-2\sin^2 \frac{1}{2}x )}{x^2. \sin x} .[/tex]
[tex]\sf = lim_{x\to 0}\ \dfrac{(x) (-2) (\frac{1}{2}x )^2}{x^2. (x)} .[/tex]
[tex]\sf = lim_{x\to 0}\ \dfrac{-2(\frac{1}{4})}{1} = - \dfrac{1}{2} .[/tex]
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