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trigonometri
Penjelasan dengan langkah-langkah:
[tex]\sf 1. lim_{x\to 0} \dfrac{\tan2x. \sin^2 8x}{x^2.\sin 4x}[/tex]
[tex]\sf lim_{x\to 0} \dfrac{2x. (8x)^2}{x^2. (4x) } = \dfrac{128 x^3}{4^3} = 32\\[/tex]
[tex]2. \lim_{x\to 0}\dfrac{1-\cos 4x}{\tan^2 3x}[/tex]
[tex]\lim_{x\to 0}\dfrac{2\sin^2 2x}{\tan^2 3x}= \dfrac{2(2x)^2}{(3x)^2} = \dfrac{8}{9}\\[/tex]
[tex]3.\sf lim_{x\to 3} \dfrac{(x+2) \tan (x-3)}{2x^2- 5x - 3}[/tex]
[tex]\sf lim_{x\to 3} \dfrac{(x+2) \tan (x-3)}{(2x+1)(x- 3)}[/tex]
[tex]\sf lim_{x\to 3} \dfrac{(x+2)}{(2x+1)}= \dfrac{3+2}{2(3) + 1} = \dfrac{5}{7}\\[/tex]
[tex]4.\sf lim_{x\to 0}\dfrac{x\tan 5x}{\cos 2x - \cos 7x}[/tex]
[tex]\sf lim_{x\to 0}\dfrac{x\tan 5x}{- (\cos 7x - \cos 2x)}[/tex]
[tex]\sf lim_{x\to 0}\dfrac{x\tan 5x}{- (- 2 \sin \frac{9}{2}x \sin \frac{5}{2}x)}[/tex]
[tex]\sf lim_{x\to 0}\dfrac{x\tan 5x}{2 \sin \frac{9}{2}x \sin \frac{5}{2}x}[/tex]
[tex]\sf lim_{x\to 0}\dfrac{x(5x)}{2 (\frac{9}{2})(\frac{5}{2}x)}=\dfrac{10}{45} = \dfrac{2}{9}\\[/tex]
[tex]5.\sf lim_{x\to 0} \dfrac{\sin 3x- \sin 3x\cos 2x}{x \tan^2 x}[/tex]
[tex].\sf lim_{x\to 0} \dfrac{\sin 3x(1- \cos 2x)}{x \tan^2 x}[/tex]
[tex].\sf lim_{x\to 0} \dfrac{\sin 3x(2\sin^2 x)}{x \tan^2 x}[/tex]
[tex].\sf lim_{x\to 0} \dfrac{3x(2x^2)}{x (x^2)} = \dfrac{6x^3}{x^3} = 6\\[/tex]
[tex]6. \sf lim_{x\to \frac{\pi}{4}} \dfrac{1- 2\sin x\cos x}{\sin x - \cos x}[/tex]
[tex]\sf lim_{x\to \frac{\pi}{4}} \dfrac{(\sin x - \cos x)(\sin x - \cos x)}{\sin x - \cos x}[/tex]
[tex]\sf lim_{x\to \frac{\pi}{4}}\ \ {\sin x - \cos x}[/tex]
[tex]\sf lim_{x\to \frac{\pi}{4}}\ \sin \frac{\pi}{4} - \cos \frac{\pi}{4}[/tex]
[tex]\sf= \frac{1}{2}\sqrt 2- \frac{1}{2}\sqrt 2 = 0[/tex]
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