nilai limid mendekati nol dari sec x + cos x -2 dibagi x kuadrat sin kuadrat x??

[tex]\displaystyle \lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\lim_{x\to0}\frac{\frac1{\cos x}+\cos x-2}{x^2\sin^2x}\\\lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\lim_{x\to0}\frac{\frac{\cos^2x-2\cos x+1}{\cos x}}{x^2\sin^2x}\\\lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\lim_{x\to0}\frac{(\cos x-1)^2}{x^2\sin^2x\cos x}\\\lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\lim_{x\to0}\frac{(-2\sin^2\frac12x)^2}{x^2\sin^2x\cos x}[/tex]
[tex]\displaystyle \lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\lim_{x\to0}\left(\frac{-2\sin^2\frac12x}{x\sin x}\right)^2\cdot\frac1{\cos x}\\\lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\left(-2\cdot\frac{1}{4}\right)^2\cdot\frac1{\cos0}\\\lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\left(-\frac12\right)^2\cdot\frac1{1}\\\boxed{\boxed{\lim_{x\to0}\frac{\sec x+\cos x-2}{x^2\sin^2x}=\frac14}}[/tex]

Lim x->0 sec x + cos x - 2 / x^2 sin^2x
Aturan L'Hopital
Lim x->0 sec x + cos x - 2 / x^2 sin^2x . x^2/x^2
Lim x->0 sec x + cos x - 2 / x^4
Lim x->0 sin x tan^2x / 4x^3
= 1/4 -> Jawab