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Question Number 125204 by bemath last updated on 09/Dec/20

Answered by liberty last updated on 09/Dec/20

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(3^{4x}\right)}^{\frac{1}{2x}}{(1+\frac{1}{3^{3x}})}^{\frac{1}{2x}}=$$$$9\times \underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{3^{3x}})}^{\frac{1}{2x}}=9\times e^{\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{3^{2x}}-1).\frac{1}{2x}}$$$$=9\times e^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{1}{2x.3^{2x}}\right)}=9\times e^0=9$$

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

$${\underset{x\to \mathrm{\infty }}{\lim 3}}^2{(1+\frac{1}{3^{3x}})}^{\frac{1}{2x}}={\underset{x\to \mathrm{\infty }}{\lim 3}}^2{(1+\frac{1}{9^x})}^{\frac{1}{2x}}=3^2(1+\frac{1}{9^x.2x}).$$$$=9$$

Answered by mathmax by abdo last updated on 09/Dec/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={(3^{\mathrm{x}}+3^{4\mathrm{x}})}^{\frac{1}{2\mathrm{x}}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\left(3^{4\mathrm{x}}\right)}^{\frac{1}{2\mathrm{x}}}{\{1+\frac{1}{3^{\mathrm{x}}}\}}^{\frac{1}{2\mathrm{x}}}$$$$=9{\mathrm{e}}^{\frac{1}{2\mathrm{x}}\ln (1+\frac{1}{3^{\mathrm{x}}})}\mathrm{but}\ln (1+\frac{1}{3^{\mathrm{x}}})\sim \frac{1}{3^{\mathrm{x}}}(\mathrm{x}\to +\mathrm{\infty })\Rightarrow$$$${\mathrm{e}}^{\frac{1}{2\mathrm{x}}\ln (1+\frac{1}{3^{\mathrm{x}}})}\sim {\mathrm{e}}^{\frac{1}{2\mathrm{x}.3^{\mathrm{x}}}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim 9{\mathrm{e}}^{\frac{1}{{2\mathrm{x}3}^{\mathrm{x}}}}\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=9$$

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