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Question Number 66084 by F_Nongue last updated on 09/Aug/19

$$Howtosolvethislimit?$$$$\underset{x\to \mathrm{\infty }}{lim}{(7x+\frac{2}{x})}^x$$

Commented by mathmax by abdo last updated on 09/Aug/19

$$letA\left(x\right)={(7x+\frac{2}{x})}^x\Rightarrow A\left(x\right)={\left(7x\right)}^x{(1+\frac{2}{7x^2})}^x$$$$={\left(7x\right)}^x{e}^{xln(1+\frac{2}{7x^2})}butln(1+\frac{2}{7x^2})\sim \frac{2}{7x^2}\Rightarrow xln(1+\frac{2}{7x^2})\sim \frac{2}{7x}\Rightarrow$$$$A\left(x\right)\sim {\left(7x\right)}^x\times \frac{2}{7x}=2{\left(7x\right)}^{x-1}=2e^{(x-1)ln\left(7x\right)}\to +\mathrm{\infty }(x\to +\mathrm{\infty })\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}{(7x+\frac{2}{x})}^x=+\mathrm{\infty }$$

Answered by MJS last updated on 09/Aug/19

$${(7x+\frac{2}{x})}^x={\left(\frac{7x^2+2}{x}\right)}^x=\frac{{(7x^2+2)}^x}{x^x}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{(7x^2+2)}^x}{x^x}>\underset{x\to \mathrm{\infty }}{\lim }\frac{{\left(7x^2\right)}^x}{x^x}={\underset{x\to \mathrm{\infty }}{\lim 7}}^x{x}^x=+\mathrm{\infty }$$

Commented by Prithwish sen last updated on 09/Aug/19

$$\frac{2}{\mathrm{x}}\to 0\mathrm{as}\mathrm{x}\to \mathrm{\infty }\therefore {(7\mathrm{x}+\frac{2}{\mathrm{x}})}^{\mathrm{x}}\to 7^{\mathrm{x}}{\mathrm{x}}^{\mathrm{x}}$$

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