LImit Trigonometri
x mendekati 0
Penjelasan dengan langkah-langkah:
[tex]\sf 1) lim_{x\to 0}\ \dfrac{x \tan 2x}{\sin^2 2x - \cos 4x + 1}[/tex]
[tex]\sf lim_{x\to 0}\ \dfrac{x \tan 2x}{\sin^2 2x - (1- 2 \sin^2 2x) + 1}[/tex]
[tex]\sf lim_{x\to 0}\ \dfrac{x \tan 2x}{\sin^2 2x - 1 +2 \sin^2 2x + 1}[/tex]
[tex]\sf lim_{x\to 0}\ \dfrac{x \tan 2x}{3\sin^2 2x }[/tex]
[tex]\sf lim_{x\to 0}\ \dfrac{x \tan 2x}{3\sin 2x . \sin 2x}[/tex]
[tex]\sf = \dfrac{x(2x)}{3(2x)(2x)} = \dfrac{2x^2}{12x^2} = \dfrac{1}{6}[/tex]
.
[tex]\sf 2). lim_{x\to 0}\ \dfrac{5 \sin^2 x - x^2\cos^2 x}{x \tan x}[/tex]
[tex]\sf lim_{x\to 0}\ \dfrac{5 \sin^2 x}{x\tan x} - lim_{x\to 0}\dfrac{x^2\cos^2 x}{x \tan x}[/tex]
[tex]\sf lim_{x\to 0}\ \dfrac{5 \sin^2 x}{x\tan x} - lim_{x\to 0}\dfrac{x^2}{x \tan x}[/tex]
[tex]\sf lim_{x\to 0}\ \dfrac{5 x^2}{x^2} - lim_{x\to 0}\dfrac{x^2}{x^2}[/tex]
= 5- 1
= 4
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