Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 84902 by john santu last updated on 17/Mar/20

$$\underset{x\to \mathrm{\infty }}{\lim }(5^{\mathrm{x}}+5^{2\mathrm{x}}){}^{\frac{1}{\mathrm{x}}}?$$

Commented by john santu last updated on 17/Mar/20

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{5^{\mathrm{x}}+5^{2\mathrm{x}}}{5^{2\mathrm{x}}}\times 5^{2\mathrm{x}}\right)}^{\frac{1}{\mathrm{x}}}=$$$$5^2\times \underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{5^{\mathrm{x}}})}^{\frac{1}{\mathrm{x}}}=$$$$25\times {\mathrm{e}}^{\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{5^{\mathrm{x}}}-1).\frac{1}{\mathrm{x}}}=$$$$25\times {\mathrm{e}}^0=25\times 1=25$$

Commented by mathmax by abdo last updated on 17/Mar/20

$$letf\left(x\right)={(5^x+5^{2x})}^{\frac{1}{x}}\Rightarrow f\left(x\right)=e^{\frac{1}{x}ln(5^{2x}+5^x)}$$$$=e^{\frac{1}{x}\left(2x\right)ln\left(5\right)+ln(1+5^{-x})}=25e^{ln(1+5^{-x})}\Rightarrow f\left(x\right)\sim 25e^{5^{-x}}\to 25(x\to +\mathrm{\infty })$$$$\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=25$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com