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

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

[tex]
\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
[/tex]