Question Number 13011 by sandy_suhendra last updated on 10/May/17
![lim_(x→∞) [5^x + 5^(3x) ]^(1/x) = ? please help](https://www.tinkutara.com/question/Q13011.png)
$$\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\left[\mathrm{5}^{\mathrm{x}} \:+\:\mathrm{5}^{\mathrm{3x}} \right]^{\frac{\mathrm{1}}{\mathrm{x}}} \:=\:? \\ $$$$\mathrm{please}\:\mathrm{help} \\ $$
Answered by sma3l2996 last updated on 10/May/17
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{5}^{{x}} +\mathrm{5}^{\mathrm{3}{x}} \right)^{\frac{\mathrm{1}}{{x}}} =\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{5}^{\mathrm{3}{x}} \left(\mathrm{5}^{−\mathrm{2}{x}} +\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{{x}}} =\underset{{x}\rightarrow+\infty} {\mathrm{lim}5}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{5}^{−\mathrm{2}{x}} \right)^{\frac{\mathrm{1}}{{x}}} \\ $$$${we}\:{have}\underset{{x}\rightarrow+\infty} {\mathrm{lim}5}^{−\mathrm{2}{x}} =\mathrm{0} \\ $$$${so}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{5}^{{x}} +\mathrm{5}^{\mathrm{3}{x}} \right)^{\frac{\mathrm{1}}{{x}}} =\underset{{x}\rightarrow+\infty} {\mathrm{lim}5}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{5}^{−\mathrm{2}{x}} \right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{5}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{0}} =\mathrm{5}^{\mathrm{3}} =\mathrm{125} \\ $$
Commented by sandy_suhendra last updated on 11/May/17
$$\mathrm{thank}'\mathrm{s}\:\mathrm{for}\:\mathrm{your}\:\mathrm{kindness} \\ $$