tolongg,yang bisa limitt makasih

Limit Trigonometri
Untuk Lim x -> 0 Gunakan Aturan L'Hopital
Lim x -> 0 (sin 2x + tan 3x) / (x cos 2x)
Lim x -> 0 (sin 2x . 2x/2x + tan 3x . 3x/3x) / (x cos 2x)
Lim x -> 0 (2x + 3x) / (x cos 2x)
Lim x -> 0 5x / x cos 2x
Lim x -> 0 5 / cos 2x
5 (E) -> Jawab

[tex]\displaystyle \lim_{x\to0}\frac{\sin2x+\tan3x}{x\cos2x}=\lim_{x\to0}\frac{\sin2x+\tan3x}{x(1-2\sin^2x)}\\\lim_{x\to0}\frac{\sin2x+\tan3x}{x\cos2x}=\lim_{x\to0}\frac{\sin2x+\tan3x}{x(1-2\sin^2x)}\cdot\frac{\frac1x}{\frac1x}\\\lim_{x\to0}\frac{\sin2x+\tan3x}{x\cos2x}=\lim_{x\to0}\frac{\frac{\sin2x}x+\frac{\tan3x}x}{1-2\sin^2x}\\\lim_{x\to0}\frac{\sin2x+\tan3x}{x\cos2x}=\frac{\lim_{x\to0}\frac{\sin2x}x+\frac{\tan3x}x}{\lim_{x\to0}1-2\sin^2x}[/tex]
[tex]\displaystyle \lim_{x\to0}\frac{\sin2x+\tan3x}{x\cos2x}=\frac{2+3}{1-2\sin^20}\\\boxed{\boxed{\lim_{x\to0}\frac{\sin2x+\tan3x}{x\cos2x}=5}}[/tex]