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[tex] \displaystyle\sf \lim_{x \to2} \: \frac{ {x}^{a} - {2}^{a} }{ {x}^{ \frac{a}{3} } - {2}^{ \frac{a}{3} } } = 6 \sqrt[3]{2} [/tex]
Tentukan nilai a !
Pilihan ganda
a. 1
b. 2
c. 3
d. 4
e. semua salah
Jawaban:
a = 2
Pembahasan
Limit
[tex]\large\text{$\begin{aligned}{\Biggl.}6\sqrt[3]{2}&=\lim_{x\to2}\:\frac{x^a-2^a}{x^{\frac{a}{3}}-2^{\frac{a}{3}}}\\{\Biggl.}\quad&\left[\ \normalsize{\begin{aligned}&\textsf{bentuk $\tfrac{0}{0}$, maka terapkan}\\&\textsf{aturan L'H\^opital}\end{aligned}}\right.\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}{\Biggl.}6\sqrt[3]{2}&=\lim_{x\to2}\:\frac{\frac{d}{dx}\left(x^a-2^a\right)}{\frac{d}{dx}\left(x^{\frac{a}{3}}-2^{\frac{a}{3}}\right)}\\{\Biggl.}&=\lim_{x\to2}\:\frac{ax^{a-1}-0}{\frac{a}{3}x^{\frac{a}{3}-1}-0}\\{\Biggl.}&=\lim_{x\to2}\:\frac{\cancel{a}x^{a-1}}{\frac{\cancel{a}}{3}x^{\frac{a}{3}-1}}\\{\Biggl.}&=\lim_{x\to2}\:3\cdot\frac{x^a\cdot \cancel{x^{-1}}}{x^{\frac{a}{3}}\cdot \cancel{x^{-1}}}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}{\Biggl.}&=3\cdot\lim_{x\to2}\:x^{a-\frac{a}{3}}\\{\Biggl.}&=3\cdot\lim_{x\to2}\:x^{\frac{2a}{3}}\\6\sqrt[3]{2}&=3\cdot 2^{\frac{2a}{3}}\\2\sqrt[3]{2}&=2^{\frac{2a}{3}}\\2^{1+\frac{1}{3}}&=2^{\frac{2a}{3}}\iff2^{\frac{4}{3}}=2^{\frac{2a}{3}}\\{\Biggl.}\frac{4}{\cancel{3}}&=\frac{2a}{\cancel{3}}\iff4=2a\\\therefore\ a&=\bf2\end{aligned}$}[/tex]
