Limit bentuk tak tentu
0/0 , ~/ ~, ~ - ~
Penjelasan dengan langkah-langkah:
[tex]1. \lim_{x\to 0} \ \dfrac{3x^2}{\sin^2 2x}[/tex]
[tex]= 3 \lim_{x\to 0} \ \dfrac{x. x }{\sin 2x. \sin 2x}[/tex]
[tex]=\frac{3(1)(1)}{2(2)} = \frac{3}{4}\\\\[/tex]
[tex]2. \lim_{x\to0}\ \dfrac{\sin^3 x}{2x^3}[/tex]
[tex]= \frac{1}{2}\ . \ \lim_{x\to0}\ \dfrac{\sin^3 x}{x^3}[/tex]
[tex]= \frac{1}{2}(1) = \frac{1}{2}\\\\[/tex]
[tex]3. \lim_{x\to 0}\ \dfrac{\sin^24x}{\tan^2 6x}[/tex]
[tex]= \lim_{x\to 0}\ \dfrac{\sin4x.\sin4x}{\tan 6x. \tan 6x}[/tex]
[tex]= \dfrac{4(4)}{6(6)} = \dfrac{4}{9}\\\\[/tex]
[tex]4. \lim_{x\to \infty} \dfrac{-4x^2 + 2x^3 - 5x^4}{x^4-3x^2 + 2x - 1}[/tex]
[tex]\sf * bagikan \ dgn \ x ^4, maka[/tex]
[tex]=\lim_{x\to \infty} \dfrac{ - 5x^4}{x^4} = - 5\\\\[/tex]
[tex]5. \lim_{x\to \infty} \sqrt{4x^2 - 4x-2} - 2x + 1\\a= 4, b = - 4, q = 1\\\lim = \frac{b}{2\sqrt a } + 9 \to = \frac{-4}{2\sqrt 4}+ 1= 0[/tex]
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