Question Number 162618 by LEKOUMA last updated on 30/Dec/21
$${Calculate} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{5}^{{x}} +\mathrm{8}^{{x}} \right)^{\frac{\mathrm{1}}{\mathrm{3}{x}}} \\ $$
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](https://www.tinkutara.com/question/Q162619.png)
$$\mathcal{A}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{5}^{{x}} +\mathrm{8}^{{x}} \right)^{\frac{\mathrm{1}}{\mathrm{3}{x}}} \\ $$$$\mathrm{ln}\mathcal{A}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{3}{x}}\mathrm{ln}\left(\mathrm{5}^{{x}} +\mathrm{8}^{{x}} \right) \\ $$$$\mathrm{ln}\mathcal{A}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{3}{x}}\mathrm{ln}\left[\mathrm{8}^{{x}} \left(\mathrm{1}+\frac{\mathrm{5}^{{x}} }{\mathrm{8}^{{x}} }\right)\right] \\ $$$$\mathrm{ln}\mathcal{A}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{3}{x}}\mathrm{ln8}^{{x}} +\frac{\mathrm{1}}{\mathrm{3}{x}}\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{5}}{\mathrm{8}}\right)^{{x}} \right)\right\} \\ $$$$\mathrm{ln}\mathcal{A}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln8}+\frac{\mathrm{1}}{\mathrm{3}{x}}\left(\frac{\mathrm{5}}{\mathrm{8}}\right)^{{x}} \right\}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln8}=\mathrm{ln2} \\ $$$$\mathcal{A}=\mathrm{2} \\ $$
Answered by tounghoungko last updated on 31/Dec/21
![lim_(x→∞) [8^x (((5/8))^x +1)]^(1/(3x)) = lim_(x→∞) 2(((5/8))^x +1)^(1/(3x)) = 2×1=2](https://www.tinkutara.com/question/Q162637.png)
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{8}^{{x}} \:\left(\left(\frac{\mathrm{5}}{\mathrm{8}}\right)^{{x}} +\mathrm{1}\right)\right]^{\frac{\mathrm{1}}{\mathrm{3}{x}}} \:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}\left(\left(\frac{\mathrm{5}}{\mathrm{8}}\right)^{{x}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}{x}}} =\:\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$