Bambang Hariyanto: Pembahasan Limit Trigonometri

SOAL :(NO TURUNAN)
$\\ \\=\lim_{x\rightarrow 0}\frac{\sqrt{2}-cosx-sinx}{(4x-\pi )^{2}} \\$
JAWAB :

$\\lim_{x\rightarrow 0}\frac{\sqrt{2}-cosx-sinx}{\left ( 4x-\pi \right )^{2}} \\ \\lim_{x\rightarrow 0}\frac{\sqrt{2}-[sin(\frac{\pi}{2}-x)+sinx]}{\left ( \pi -4x \right )^{2}} \\ \\lim_{x\rightarrow 0}\frac{\sqrt{2}-[2sin\frac{1}{2}(\frac{\pi}{2}-x+x)cos\frac{1}{2}(\frac{\pi}{2}-x-x)]}{16\left ( \frac{\pi }{4} -x \right )^{2}} \\ \\lim_{x\rightarrow 0}\frac{\sqrt{2}-[2.\frac{1}{2}\sqrt{2}.cos(\frac{\pi}{4}-x)]}{16\left ( \frac{\pi }{4} -x \right )^{2}} \\ \\lim_{x\rightarrow 0}\frac{\sqrt{2}-\sqrt{2}.cos(\frac{\pi}{4}-x)}{16\left ( \frac{\pi }{4} -x \right )^{2}} \\ \\lim_{x\rightarrow 0}\frac{\sqrt{2}(1-cos(\frac{\pi}{4}-x))}{16\left ( \frac{\pi }{4} -x \right )^{2}} \\ \\lim_{x\rightarrow 0}\frac{\sqrt{2}.2sin\frac{1}{2}(\frac{\pi}{4}-x).sin\frac{1}{2}(\frac{\pi}{4}-x)}{16\left ( \frac{\pi }{4} -x \right )(\frac{\pi }{4}-x)} \\$
$\\ \\=\frac{\sqrt{2}.2.\frac{1}{2}.\frac{1}{2}}{16} \\ \\=\frac{\sqrt{2}}{32} \\$

SEMOGA BERMANFAAT

Bambang Hariyanto

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Tuesday, 21 August 2018

Pembahasan Limit Trigonometri

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