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Question Number 120201 by bemath last updated on 30/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (x+e^x+e^{2x})}{x}=?$$

Answered by benjo_mathlover last updated on 30/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (x+e^x+e^{2x})}{x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1+e^x+2e^{2x}}{x+e^x+e^{2x}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{e^{2x}(e^{-2x}+e^{-x}+2)}{e^{2x}({xe}^{-2x}+e^{-x}+1)}=\frac{0+0+2}{0+0+1}=2$$

Answered by Dwaipayan Shikari last updated on 30/Oct/20

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

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