Berikut ini adalah kumpulan soal dan pembahasan mengenai limit tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 11
{\displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x^{\frac{n}3}-2^{\frac{n}3}}=3\sqrt[3]{16}}, tentukan nilai
n
CARA 1 : PEMFAKTORAN
\begin{aligned}
\displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x^{\frac{n}3}-2^{\frac{n}3}}&=3\sqrt[3]{16}\\
\displaystyle\lim_{x\to2}\dfrac{\left(x^{\frac{n}3}\right)^3-\left(2^{\frac{n}3}\right)^3}{x^{\frac{n}3}-2^{\frac{n}3}}&=3\sqrt[3]{16}\\
\displaystyle\lim_{x\to2}\dfrac{\left(x^{\frac{n}3}-2^{\frac{n}3}\right)\left(x^{\frac{2n}3}+x^{\frac{n}3}2^{\frac{n}3}+2^{\frac{2n}3}\right)}{x^{\frac{n}3}-2^{\frac{n}3}}&=3\sqrt[3]{16}\\
\displaystyle\lim_{x\to2}\left(x^{\frac{2n}3}+x^{\frac{n}3}2^{\frac{n}3}+2^{\frac{2n}3}\right)&=3\sqrt[3]{16}\\
2^{\frac{2n}3}+2^{\frac{n}3}2^{\frac{n}3}+2^{\frac{2n}3}&=3\sqrt[3]{16}\\
2^{\frac{2n}3}+2^{\frac{2n}3}+2^{\frac{2n}3}&=3\sqrt[3]{16}\\
3\cdot2^{\frac{2n}3}&=3\sqrt[3]{16}\\
3\sqrt[3]{2^{2n}}&=3\sqrt[3]{16}\\
2^{2n}&=16\\
2n&=4\\
n&=\boxed{\boxed{2}}
\end{aligned}
CARA 2 : L'HOPITAL
\begin{aligned}
\displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x^{\frac{n}3}-2^{\frac{n}3}}&=3\sqrt[3]{16}\\
\displaystyle\lim_{x\to2}\dfrac{nx^{n-1}}{\dfrac{n}3x^{\frac{n}3-1}}&=3\sqrt[3]{16}\\
\displaystyle\lim_{x\to2}3x^{n-\frac{n}3}&=3\sqrt[3]{16}\\
\displaystyle\lim_{x\to2}3x^{\frac{2n}3}&=3\sqrt[3]{16}\\
\displaystyle\lim_{x\to2}3\sqrt[3]{x^{2n}}&=3\sqrt[3]{16}\\
3\sqrt[3]{2^{2n}}&=3\sqrt[3]{16}\\
2^{2n}&=16\\
2n&=4\\
n&=\boxed{\boxed{2}}
\end{aligned}
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