Tolong bantuannya cepat ya.... Question from @veraards

whongaliem $\lim_{x \to \i0} \frac{sin 4x + sin 4x . cos 2x}{4 x^{3} }$ =
$\lim_{x \to \i0} \frac{sin 4x (1 - cos 2x) }{4 x^{3} }$ =
$\lim_{x \to \i0} \frac{sin 4x (1 - cos^{2}x + sin^{2}x ) }{4. x^{3} }$ =
$\lim_{x \to \i0} \frac{sin 4x ( sin^{2}x + sin^{2}x) }{4. x^{3} }$ =
$\lim_{x \to \i0} \frac{sin 4x (2 sin^{2}x )}{4. x^{3} }$ =
$\lim_{x \to \i0} \frac{2}{4}. \frac{sin 4x}{x}. \frac{sin x}{x} \frac{sin x}{x}$ =
$\frac{1}{2}$ . 4 . 1 . 1 = 2
-) $\lim_{x \to \infty} \sqrt{4 x^{2} + 8x - 9} - (2x + 4)$ ---- kalikan denan sekawannya
$\lim_{x \to \infty} \frac{( 4x^{2} + 8x - 9) - (x + 4)^{2} }{ (\sqrt{ 4x^{2}+ 8x - 9 })+ (x + 4 )}$ =
$\lim_{x \to \infty} \frac{ 4x^{2}+ 8x - 9 - 4x^{2}- 16x + 16 }{2x. \sqrt{1 + \frac{8}{x}- \frac{9}{ x^{2} } }+ 2x(1 + \frac{2}{x}) }$ =
$\lim_{x \to \infty} \frac{- 8x + 7}{(2x \sqrt{1 + \frac{4}{x}- \frac{9}{ x^{2} })+ 2x (1 + \frac{2}{x} ) } }$ = ....bagi dengan x ----> x/∞ = 0
$\frac{-8}{2 + 2}$ =
$\frac{- 8}{4}$ = 2

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