1.limit x mendekati 0 untuk x - sin 2x / 2x + Tan x =… 2.limit x mendekati phi untuk 1 + cos x / sin.... Question from @KimSquirrel

April 2019 2 9 Report

1.limit x mendekati 0 untuk x - sin 2x / 2x + Tan x =…
2.limit x mendekati phi untuk 1 + cos x / sin 2x =…

To9ha Jawab

limit (x - sin 2x) / (2x + tan x)
x→0

limit (x/2x - sin 2x/2x) / (2x/2x + tan x/2x)
x → 0
= (1/2 - 1) / (1+ 1/2)
= (1 -2)/(2 +1)
= -1/3

2.
limit (1+ cos x) / (sin 2x)
x→ π
= ( - sin x) / (2 cos 2x)
= ( - sin π )/ (2 cos 2π)
= 0

6 votes Thanks 4

KimSquirrel Kenapa yang nomor 1 dibagi 2x semua?

supaat69 $\text{1. } \text{Menggunakan Dalil l'Hopital}\\ \lim_{x\to 0}{\frac{x-\sin{2x}}{2x+\tan{x}}}\\ =\lim_{x\to 0}{\frac{1-2\cos{2x}}{2+\sec^2{x}}}\\ =\frac{1-2\cos{(2\cdot0)}}{2+\sec^2{0}}\\ =\frac{1-2}{2+1}\\ =-\frac{1}{3}\\ \\ \text{2. } \text{Mengalikan dengan bentuk sekawan}\\\\\lim_{x\to\pi}{\frac{1+\cos{x}}{\sin{2x}}\cdot\frac{1-\cos{x}}{1-\cos{x}}}\\ =\lim_{x\to\pi}{\frac{1-\cos^2{x}}{\sin{2x}\cdot(1-\cos{x})}}\\ =\lim_{x\to\pi}{\frac{\sin^2{x}}{2\sin{x}\cos{x}\cdot(1-\cos{x})}}\\ =\lim_{x\to\pi}{\frac{\sin{x}}{2\cos{x}\cdot(1-\cos{x})}}\\ =\frac{\sin{\pi}}{2\cos{\pi}\cdot(1-\cos{\pi})}\\ =\frac{0}{2\cdot(-1)\cdot(1-(-1))}\\ =\frac{0}{-4}\\ =0$

0 votes Thanks 1

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