Question-170767

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

Answered by aleks041103 last updated on 30/May/22

L=lim_(x→∞) ((1/(5^x +7^x +9^x )))^(−(√(1/x))) ⇒lnL=lim_(x→∞) ((ln(5^x +7^x +9^x ))/( (√x))) since 5,7,9>1⇒5^x +7^x +9^x >9^x ⇒((ln(5^x +7^x +9^x ))/( (√x)))>((x ln(9))/( (√x)))=(√x)ln(9) ⇒since (√x) ln(9) diverges as x→∞, then ((ln(5^x +7^x +9^x ))/( (√x)))→∞ , too. ⇒L→∞

$${L}=\underset{{x}\rightarrow\infty} {{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{5}^{{x}} +\mathrm{7}^{{x}} +\mathrm{9}^{{x}} }\right)^{−\sqrt{\frac{\mathrm{1}}{{x}}}} \\ $$$$\Rightarrow{lnL}=\underset{{x}\rightarrow\infty} {{lim}}\frac{{ln}\left(\mathrm{5}^{{x}} +\mathrm{7}^{{x}} +\mathrm{9}^{{x}} \right)}{\:\sqrt{{x}}} \\ $$$${since}\:\mathrm{5},\mathrm{7},\mathrm{9}>\mathrm{1}\Rightarrow\mathrm{5}^{{x}} +\mathrm{7}^{{x}} +\mathrm{9}^{{x}} >\mathrm{9}^{{x}} \\ $$$$\Rightarrow\frac{{ln}\left(\mathrm{5}^{{x}} +\mathrm{7}^{{x}} +\mathrm{9}^{{x}} \right)}{\:\sqrt{{x}}}>\frac{{x}\:{ln}\left(\mathrm{9}\right)}{\:\sqrt{{x}}}=\sqrt{{x}}{ln}\left(\mathrm{9}\right) \\ $$$$\Rightarrow{since}\:\sqrt{{x}}\:{ln}\left(\mathrm{9}\right)\:{diverges}\:{as}\:{x}\rightarrow\infty, \\ $$$${then}\:\frac{{ln}\left(\mathrm{5}^{{x}} +\mathrm{7}^{{x}} +\mathrm{9}^{{x}} \right)}{\:\sqrt{{x}}}\rightarrow\infty\:,\:{too}. \\ $$$$\Rightarrow{L}\rightarrow\infty \\ $$

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 }) →+∞

$${A}_{{n}} ={e}^{−\frac{\mathrm{1}}{\:\sqrt{{n}}}{ln}\left(\frac{\mathrm{1}}{\mathrm{5}^{{n}} +\mathrm{7}^{{n}} +\mathrm{9}^{{n}} }\right)} \\ $$$$={e}^{\frac{\mathrm{1}}{\:\sqrt{{n}}}{ln}\left(\mathrm{5}^{{n}} +\mathrm{7}^{{n}} +\mathrm{9}^{{n}} \right)} \\ $$$$={e}^{\frac{\mathrm{1}}{\:\sqrt{{n}}}\left(\mathrm{2}{nln}\left(\mathrm{3}\right)+{ln}\left(\mathrm{1}+\left(\frac{\mathrm{5}}{\mathrm{9}}\right)^{{n}} +\left(\frac{\mathrm{7}}{\mathrm{9}}\right)^{{n}} \right)\right.} \\ $$$$={e}^{\mathrm{2}\sqrt{{n}}{ln}\left(\mathrm{3}\right)} .{e}^{\frac{\mathrm{1}}{\:\sqrt{{n}}}{ln}\left(\mathrm{1}+\left(\frac{\mathrm{5}}{\mathrm{9}}\right)^{{n}} +\left(\frac{\mathrm{7}}{\mathrm{9}}\right)^{{n}} \right)} \\ $$$$\sim{e}^{\mathrm{2}\sqrt{{n}}{ln}\left(\mathrm{3}\right)} .{e}^{\frac{\mathrm{1}}{\:\sqrt{{n}}}\left\{\left(\frac{\mathrm{5}}{\mathrm{9}}\right)^{{n}} +\left(\frac{\mathrm{7}}{\mathrm{9}}\right)^{{n}} \right\}} \rightarrow+\infty \\ $$

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