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Question Number 170767 by 2407 last updated on 30/May/22

	Answered by aleks041103 last updated on 30/May/22

	$L = \underset{x \to \infty}{l i m} {(\frac{1}{5^{x} + 7^{x} + 9^{x}})}^{- \sqrt{\frac{1}{x}}}$ $\Rightarrow l n L = \underset{x \to \infty}{l i m} \frac{l n (5^{x} + 7^{x} + 9^{x})}{\sqrt{x}}$ $s i n c e 5, 7, 9 > 1 \Rightarrow 5^{x} + 7^{x} + 9^{x} > 9^{x}$ $\Rightarrow \frac{l n (5^{x} + 7^{x} + 9^{x})}{\sqrt{x}} > \frac{x l n (9)}{\sqrt{x}} = \sqrt{x} l n (9)$ $\Rightarrow s i n c e \sqrt{x} l n (9) d i v e r g e s a s x \to \infty,$ $t h e n \frac{l n (5^{x} + 7^{x} + 9^{x})}{\sqrt{x}} \to \infty, t o o .$ $\Rightarrow L \to \infty$
	Answered by Mathspace last updated on 31/May/22
	$A_n =e^(−(1/( (√n)))ln((1/(5^n +7^n +9^n )))) =e^((1/( (√n)))ln(5^n +7^n +9^n )) =e^((1/( (√n)))(2nln(3)+ln(1+((5/9))^n +((7/9))^n )) =e^(2(√n)ln(3)) .e^((1/( (√n)))ln(1+((5/9))^n +((7/9))^n )) ∼e^(2(√n)ln(3)) .e^((1/( (√n))){((5/9))^n +((7/9))^n }) →+∞$
	$A_{n} = e^{- \frac{1}{\sqrt{n}} l n (\frac{1}{5^{n} + 7^{n} + 9^{n}})}$ $= e^{\frac{1}{\sqrt{n}} l n (5^{n} + 7^{n} + 9^{n})}$ $= e^{\frac{1}{\sqrt{n}} (2 n l n (3) + l n (1 + {(\frac{5}{9})}^{n} + {(\frac{7}{9})}^{n})}$ $= e^{2 \sqrt{n} l n (3)} . e^{\frac{1}{\sqrt{n}} l n (1 + {(\frac{5}{9})}^{n} + {(\frac{7}{9})}^{n})}$ $\sim e^{2 \sqrt{n} l n (3)} . e^{\frac{1}{\sqrt{n}} {{(\frac{5}{9})}^{n} + {(\frac{7}{9})}^{n}}} \to + \infty$
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