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[d1z4ft]

$\lim_{x \to \infty}{(8^{n}+4^{n})}^{\frac{1}{3}}-2^{n}$
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ข้อความเดิมเขียนโดยคุณ d1z4ft

$\lim_{x \to \infty}{(8^{n}+4^{n})}^{\frac{1}{3}}-2^{n}$
ทำไงงับ

$\lim_{x \to \infty}({(8^{n}+4^{n})}^{\frac{1}{3}}-2^{n})$

$\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}}$