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Question Number 22365 by A1B1C1D1 last updated on 16/Oct/17

Answered by sma3l2996 last updated on 16/Oct/17

$$=\underset{x\to \mathrm{\infty }}{lim}\frac{1-{\left(\frac{3}{5}\right)}^x+\frac{1}{5^x}}{1+{\left(\frac{3}{5}\right)}^x+\frac{1}{5^{x}x}}=\frac{1-0+0}{1+0+0}=1$$

Commented by A1B1C1D1 last updated on 16/Oct/17

$$\mathrm{Sorry},\mathrm{but}\mathrm{which}\mathrm{method}\mathrm{did}\mathrm{you}\mathrm{use}?$$

Commented by sma3l2996 last updated on 16/Oct/17

$$\underset{x\to \mathrm{\infty }}{lim}\frac{5^x-3^x+1}{5^x+3^x+\frac{1}{x}}=\underset{x\to \mathrm{\infty }}{lim}\frac{5^x(1-\frac{3^x}{5^x}+\frac{1}{5^x})}{5^x(1+\frac{3^x}{5^x}+\frac{1}{5^x\times x})}$$$$=\underset{x\to \mathrm{\infty }}{lim}\frac{1-{\left(\frac{3}{5}\right)}^x+\frac{1}{5^x}}{1+{\left(\frac{3}{5}\right)}^x+\frac{1}{5^x\times x}}=\frac{1-{\left(\frac{3}{5}\right)}^{\mathrm{\infty }}+5^{-\mathrm{\infty }}}{1+{\left(\frac{3}{5}\right)}^{\mathrm{\infty }}+\frac{1}{\mathrm{\infty }\times e^{\mathrm{\infty }}}}=\frac{1-0+0}{1+0+0}=1$$$$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{3}{5}\right)}^x=0because\frac{3}{5}<1$$$$and\underset{x\to \mathrm{\infty }}{lim}{5}^{x}x={\underset{x\to \mathrm{\infty }}{limxe}}^{xln5}=\mathrm{\infty }$$

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