help me please, nanti tak gojekin nasi padang

Berikut ini adalah pertanyaan dari davamwehehehe pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Help me please, nanti tak gojekin nasi padang

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Nomor 1

$\large\text{$\begin{aligned}\lim_{x\to 0}\:\frac{1-\cos8x}{2x\tan4x}=\boxed{\,\bf4\,}\end{aligned}$}$

Nomor 2

$\large\text{$\begin{aligned}&\lim_{x\to4}\:\frac{x^2-1}{x^2-2x-8}=\sf\underline{divergen}\\&\Rightarrow \textsf{Nilai limitnya \underline{tidak ada}.}\end{aligned}$}$
 

Pembahasan

Menentukan Nilai Limit

Nomor 1

$\begin{aligned}&\lim_{x\to 0}\:\frac{1-\cos8x}{2x\tan4x}\\&{=\ }\lim_{x\to 0}\left(\frac{1}{2}\cdot\frac{1-\cos8x}{x\tan4x}\right)\\&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{1-\cos8x}{x\tan4x}\\&\quad\left[\ \begin{aligned}&\tan\left(\tfrac{1}{2}\alpha\right)=\frac{\sin\alpha}{1+\cos\alpha}\\&\Rightarrow \tan4x=\frac{\sin8x}{1+\cos8x}\end{aligned}\right.\end{aligned}$
$\begin{aligned}&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{1-\cos8x}{x\left(\dfrac{\sin8x}{1+\cos8x}\right)}\\&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{(1-\cos8x)(1+\cos8x)}{x\sin8x}\\&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{1-\cos^28x}{x\sin8x}\\&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{\sin^28x}{x\sin8x}\\&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{\sin8x}{x}\\&\quad\to\textsf{Limit masih berbentuk 0/0.}\\&\quad\to\textsf{Memenuhi syarat aturan L'H\^{o}pital.}\end{aligned}$
$\begin{aligned}&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{\frac{d}{dx}(\sin8x)}{\frac{d}{dx}(x)}\\&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:\frac{8\cos8x}{1}\\&{=\ }\frac{1}{2}\cdot\lim_{x\to 0}\:8\cos8x\\&{=\ }\frac{1}{2}\cdot8\cos(8\cdot0)\\&{=\ }\frac{1}{2}\cdot8\\&{=\ }\boxed{\,\bf4\,}\end{aligned}$
$\blacksquare$

Nomor 2

$\begin{aligned}&\lim_{x\to4}\:\frac{x^2-1}{x^2-2x-8}\\&\quad\to\textsf{Limit bentuk tak tentu,}\\&\quad\to\textsf{namun bukan 0/0 atau $\infty/\infty$.}\\&\quad\to\textsf{Syarat aturan L'H\^opital tak terpenuhi.}\\&\quad\to\textsf{Coba faktorkan.}\end{aligned}$
$\begin{aligned}&{=\ }\lim_{x\to4}\:\frac{(x+1)(x-1)}{(x+2)(x-4)}\\&{=\ }\lim_{x\to4}\:\left(\frac{(x+1)(x-1)}{x+2}\cdot\frac{1}{x-4}\right)\\&{=\ }\lim_{x\to4}\:\frac{(x+1)(x-1)}{x+2}\cdot\lim_{x\to4}\:\frac{1}{x-4}\\&{=\ }\frac{15}{6}\cdot\lim_{x\to4}\:\frac{1}{x-4}\end{aligned}$
$\begin{aligned}.&\quad\left[\ \begin{aligned}&\bullet\ \textsf{Nilai limit dari kiri $x=4$:}\\&\quad\lim_{x\to4-}\:\frac{1}{x-4}=-\infty\\&\bullet\ \textsf{Nilai limit dari kanan $x=4$:}\\&\quad\lim_{x\to4+}\:\frac{1}{x-4}=+\infty\\&\Rightarrow\lim_{x\to4-}\:\frac{1}{x-4}\ne\lim_{x\to4+}\:\frac{1}{x-4}\\&\Rightarrow\lim_{x\to4}\frac{1}{x-4}\sf\ \;\underline{divergen},\ atau\\&\quad\textsf{nilai $\lim_{x\to4}\frac{1}{x-4}$ \underline{tidak ada}.}\end{aligned}\right.\end{aligned}$
$\begin{aligned}&{=\ }\frac{15}{6}\cdot(\sf divergen)\\&{=\ }\sf\underline{divergen}\ !\end{aligned}$
$\blacksquare$