อ้างอิง:
ข้อความเดิมเขียนโดยคุณ d1z4ft
$\lim_{x \to \infty}{(8^{n}+4^{n})}^{\frac{1}{3}}-2^{n}$
ทำไงงับ
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$\lim_{x \to \infty}({(8^{n}+4^{n})}^{\frac{1}{3}}-2^{n})$
$\qquad = \lim_{x \to \infty}({(8^{n}+4^{n})}^{\frac{1}{3}}-2^{n})
\left(\,\frac{{{(8^{n}+4^{n})}^{\frac{2}{3}}+{(8^{n}+4^{n})}^{\frac{1}{3}}2^{n}+2^{2n}}}{{{(8^{n}+4^{n})}^{\frac{2}{3}}+{(8^{n}+ 4^{n})}^{\frac{1}{3}}2^{n}+2^{2n}}}\right) $
$ \qquad = \lim_{x \to \infty}(\frac{{(8^{n}+4^{n})}-2^{3n}}{{(8^{n}+4^{n})}^{\frac{2}{3}}+{(8^{n}+4^{n})}^{\frac{1}{3}}2^{n}+2^{2n}} )$
$ \qquad = \lim_{x \to \infty}(\frac{{4^{n}}}{{(8^{n}+4^{n})}^{\frac{2}{3}}+{(8^{n}+4^{n})}^{\frac{1}{3}}2^{n}+2^{2n}} )$
$ \qquad = {\frac{1}{3}}$