Question Number 113438 by bemath last updated on 13/Sep/20
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}\right)\:? \\ $$
Answered by bemath last updated on 13/Sep/20
Answered by JDamian last updated on 13/Sep/20
![5^(1/x) =e^((ln(5))/x) =1+ (((ln(5))/x)/(1!)) + (((((ln(5))/x))^2 )/(2!)) + (((((ln(5))/x))^3 )/(3!)) + ∙∙∙ lim_(x→∞) x(5^(1/x) −1)= lim_(x→∞) (((ln(5))/(1!)) + ((([ln(5)]^2 )/x)/(2!)) + ((([ln(5)]^3 )/x^2 )/(3!)) + ∙∙∙) = ln(5).](https://www.tinkutara.com/question/Q113471.png)
$$\mathrm{5}^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{ln}\left(\mathrm{5}\right)}{{x}}} =\mathrm{1}+\:\frac{\frac{\mathrm{ln}\left(\mathrm{5}\right)}{{x}}}{\mathrm{1}!}\:+\:\frac{\left(\frac{\mathrm{ln}\left(\mathrm{5}\right)}{{x}}\right)^{\mathrm{2}} }{\mathrm{2}!}\:+\:\frac{\left(\frac{\mathrm{ln}\left(\mathrm{5}\right)}{{x}}\right)^{\mathrm{3}} }{\mathrm{3}!}\:+\:\centerdot\centerdot\centerdot \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}\right)= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{ln}\left(\mathrm{5}\right)}{\mathrm{1}!}\:+\:\frac{\frac{\left[\mathrm{ln}\left(\mathrm{5}\right)\right]^{\mathrm{2}} }{{x}}}{\mathrm{2}!}\:+\:\frac{\frac{\left[\mathrm{ln}\left(\mathrm{5}\right)\right]^{\mathrm{3}} }{{x}^{\mathrm{2}} }}{\mathrm{3}!}\:+\:\centerdot\centerdot\centerdot\right)\:=\:\boldsymbol{\mathrm{ln}}\left(\mathrm{5}\right). \\ $$