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Question Number 107207 by bemath last updated on 09/Aug/20

$$⊚bemath⊚$$$$\underset{x\to \mathrm{\infty }}{\lim }{(2^x+3^x)}^{\frac{1}{x}}?$$

Answered by Dwaipayan Shikari last updated on 09/Aug/20

$$\underset{x\to \mathrm{\infty }}{\lim 3}{(1+{\left(\frac{2}{3}\right)}^{\mathrm{x}})}^{\frac{1}{\mathrm{x}}}=\underset{x\to \mathrm{\infty }}{\lim 3}{(1+{\left(\frac{2}{3}\right)}^{\mathrm{x}})}^{\frac{1}{\mathrm{x}}{\left(\frac{3}{2}\right)}^{\mathrm{x}}{\left(\frac{2}{3}\right)}^{\mathrm{x}}}={\underset{x\to \mathrm{\infty }}{\lim 3e}}^{\frac{1}{\mathrm{x}}.{\left(\frac{2}{3}\right)}^{\mathrm{x}}}={3\mathrm{e}}^0=3$$

Answered by bemath last updated on 09/Aug/20

Answered by Dwaipayan Shikari last updated on 09/Aug/20

$$\underset{x\to \mathrm{\infty }}{\lim }{(2^{\mathrm{x}}+3^{\mathrm{x}})}^{\frac{1}{\mathrm{x}}}=\mathrm{y}$$$$\frac{1}{\mathrm{x}}\log (2^{\mathrm{x}}+3^{\mathrm{x}})=\mathrm{logy}$$$$\frac{1}{\mathrm{x}}({\log 3}^{\mathrm{x}}+\log (1+{\left(\frac{2}{3}\right)}^{\mathrm{x}}))=\mathrm{logy}$$$$\log 3+\log (1+{\left(\frac{2}{3}\right)}^{\mathrm{x}})=\mathrm{logy}$$$$\log 3=\mathrm{logy}{\left(\frac{2}{3}\right)}^{\mathrm{x}}\to 0(\mathrm{x}\to \mathrm{\infty })$$$$\mathrm{y}=3$$

Answered by 1549442205PVT last updated on 10/Aug/20

$$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{(2^x+3^x)}^{\frac{1}{x}}=\lim [3^{\mathrm{x}}{(1+{\left(\frac{2}{3}\right)}^{\mathrm{x}}]}^{\frac{1}{\mathrm{x}}}$$$$=\mathrm{l}\underset{\mathrm{x}\to \mathrm{\infty }}{\mathrm{im}}3.[{(1+{\left(\frac{2}{3}\right)}^{\mathrm{x}}]}^{\frac{1}{\mathrm{x}}}=$$$$3.\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }[{(1+{\left(\frac{2}{3}\right)}^{\mathrm{x}}]}^{\frac{1}{\mathrm{x}}}=3.{[1+{\left(\frac{2}{3}\right)}^{\mathrm{\infty }}]}^{\frac{1}{\mathrm{\infty }}}$$$$=3.{(1+0)}^0=3.1=3$$

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