Membahas Batas Fungsi \( \lim _{x \rightarrow 0} \frac{\cos 5 x-\cos 3 x}{2 \tan ^{2} x} \)

Dalam matematika, batas fungsi adalah konsep penting yang digunakan untuk memahami perilaku fungsi saat variabel mendekati suatu nilai tertentu. Dalam artikel ini, kita akan membahas batas fungsi dari ekspresi \( \lim _{x \rightarrow 0} \frac{\cos 5 x-\cos 3 x}{2 \tan ^{2} x} \) saat \( x \) mendekati nol. Pertama-tama, mari kita perjelas ekspresi tersebut. Pada pembilang, kita memiliki perbedaan antara fungsi kosinus dari \( 5x \) dan \( 3x \). Pada penyebut, kita memiliki dua kali kuadrat dari fungsi tangen dari \( x \). Untuk menentukan batas fungsi ini, kita dapat menggunakan beberapa teknik aljabar dan trigonometri. Pertama, kita dapat menggunakan identitas trigonometri \( \cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \) untuk menyederhanakan pembilang menjadi \( -2 \sin \left(\frac{8x}{2}\right) \sin \left(\frac{2x}{2}\right) \). Selanjutnya, kita dapat menggunakan identitas trigonometri \( \tan x = \frac{\sin x}{\cos x} \) untuk menyederhanakan penyebut menjadi \( 2 \left(\frac{\sin x}{\cos x}\right)^2 \). Dengan menyederhanakan ekspresi tersebut, kita mendapatkan \( \lim _{x \rightarrow 0} \frac{-2 \sin 4x \sin x}{2 \left(\frac{\sin x}{\cos x}\right)^2} \). Selanjutnya, kita dapat membagi pembilang dan penyebut dengan \( \sin x \) untuk mendapatkan \( \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\left(\frac{\sin x}{\cos x}\right)^2} \). Kemudian, kita dapat menggunakan identitas trigonometri \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \) untuk menyederhanakan penyebut menjadi \( \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\left(\frac{\sin x}{\cos x}\right)^2} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\left(\frac{\sin x}{x}\right)^2 \cos^2 x} \). Dengan menggunakan identitas trigonometri \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \) lagi, kita dapat menyederhanakan penyebut menjadi \( \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1^2 \cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} \). Selanjutnya, kita dapat menggunakan identitas trigonometri \( \cos^2 x = 1 - \sin^2 x \) untuk menyederhanakan penyebut menjadi \( \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} \). Terakhir, kita dapat menggunakan identitas trigonometri \( \sin 2x = 2 \sin x \cos x \) untuk menyederhanakan pembilang menjadi \( \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{1 - \sin^2 x} = \lim _{x \rightarrow 0} \frac{-2 \sin 4x}{\cos^2 x} = \lim