LIMit bentuk 0/0
trigonometri
aljabr
Penjelasan dengan langkah-langkah:
5. hasil dari
[tex]\sf lim_{x\to \frac{\pi}{4}}\ \dfrac{1- \tan x}{3\cos 2x}[/tex]
[tex]\sf lim_{x\to \frac{\pi}{4}}\ \dfrac{\cos x- \sin x}{3(\cos x)(\cos x -\sin x)(\cos x+ \sin x)}[/tex]
[tex]\sf lim_{x\to \frac{\pi}{4}}\ \dfrac{1}{3(\cos x)(\cos x+ \sin x)}[/tex]
cos π/4= sin π/4 = 1/2 √2
[tex]\sf lim_{x\to \frac{\pi}{4}}\ \dfrac{1}{3(\frac{1}{2}\sqrt 2)(\frac{1}{2}\sqrt 2)+(\frac{1}{2}\sqrt 2)}[/tex]
[tex]\sf lim_{x\to \frac{\pi}{4}}\ \dfrac{1}{3(\frac{1}{4}\sqrt 2)} = \frac{2}{3}\sqrt 2[/tex]
6. hasil dari
[tex]\sf lim_{x\to 3}\ \dfrac{x.\sin (2x-6)}{x^2 + 2x- 15}[/tex]
[tex]\sf lim_{x\to 3}\ \dfrac{x.\sin2(x- 3)}{(x+5)(x - 3)}[/tex]
[tex]\sf lim_{x\to 3}\ \dfrac{x}{x+5}. \ \lim_{x\to 3 }\ \dfrac{\sin2(x- 3)}{(x - 3)}[/tex]
[tex]\sf lim_{x\to 3}\ \left(\dfrac{3}{3+5}\right).\left( \ \lim_{x\to 3 }\ 2\right)[/tex]
[tex]\sf =\dfrac{3(2)}{8} = \dfrac{3}{4}[/tex]
