Calculate lim_(x→∞) (5^x +8^x )^(1/(3x))


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Question Number 162618 by LEKOUMA last updated on 30/Dec/21

$C a l c u l a t e$ $\lim_{x \to \infty} {(5^{x} + 8^{x})}^{\frac{1}{3 x}}$
	Answered by Ar Brandon last updated on 30/Dec/21
	$A=lim_(x→∞) (5^x +8^x )^(1/(3x)) lnA=lim_(x→∞) (1/(3x))ln(5^x +8^x ) lnA=lim_(x→∞) (1/(3x))ln[8^x (1+(5^x /8^x ))] lnA=lim_(x→∞) {(1/(3x))ln8^x +(1/(3x))ln(1+((5/8))^x )} lnA=lim_(x→∞) {(1/3)ln8+(1/(3x))((5/8))^x }=(1/3)ln8=ln2 A=2$
	$A = \lim_{x \to \infty} {(5^{x} + 8^{x})}^{\frac{1}{3 x}}$ $\ln A = \lim_{x \to \infty} \frac{1}{3 x} \ln (5^{x} + 8^{x})$ $\ln A = \lim_{x \to \infty} \frac{1}{3 x} \ln [8^{x} (1 + \frac{5^{x}}{8^{x}})]$ $\ln A = \lim_{x \to \infty} {\frac{1}{3 x} {\ln 8}^{x} + \frac{1}{3 x} \ln (1 + {(\frac{5}{8})}^{x})}$ $\ln A = \lim_{x \to \infty} {\frac{1}{3} \ln 8 + \frac{1}{3 x} {(\frac{5}{8})}^{x}} = \frac{1}{3} \ln 8 = \ln 2$ $A = 2$
	Answered by tounghoungko last updated on 31/Dec/21

	$\lim_{x \to \infty} {[8^{x} ({(\frac{5}{8})}^{x} + 1)]}^{\frac{1}{3 x}} =$ $\lim_{x \to \infty} 2 {({(\frac{5}{8})}^{x} + 1)}^{\frac{1}{3 x}} = 2 \times 1 = 2$
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