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Question Number 117688 by Khalmohmmad last updated on 13/Oct/20

Commented by prakash jain last updated on 13/Oct/20

$$1.\frac{exponential}{polynomial}$$$$2.\frac{polynomial}{exponential}$$

Commented by Khalmohmmad last updated on 13/Oct/20

$$\mathrm{with}\mathrm{is}\mathrm{the}\mathrm{Answer}$$

Commented by JDamian last updated on 13/Oct/20

You should know that, when base is greater than 1, the exponential function always wins any polynomial.

Answered by 1549442205PVT last updated on 13/Oct/20

$$\mathrm{a})\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{4^{\mathrm{x}}+2\mathrm{x}+3}{{\mathrm{x}}^2+6\mathrm{x}+7}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{1+\mathrm{x}.\ln 4+\frac{{\mathrm{x}}^2{\ln }^{2}4}{2}+\frac{{\mathrm{x}}^3{\ln }^{3}4}{6}++2\mathrm{x}+3}{{\mathrm{x}}^2+6\mathrm{x}+7}$$$$\underset{\mathrm{x}\to \mathrm{\infty }}{=\lim }\frac{{\mathrm{x}}^3{\ln }^{3}4}{{6\mathrm{x}}^2}=\mathrm{\infty }$$$$\mathrm{b})\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{{\mathrm{x}}^2+2\mathrm{x}+3}{3^{\mathrm{x}}+3\mathrm{x}+1}=\lim \frac{{\mathrm{x}}^2+2\mathrm{x}+3}{1+\mathrm{x}.\ln 3+\frac{{\mathrm{x}}^2{\ln }^{2}3}{2}+\frac{{\mathrm{x}}^3{\ln }^{3}3}{6}+3\mathrm{x}+1}$$$$\underset{\mathrm{x}\to \mathrm{\infty }}{=\lim }\frac{{6\mathrm{x}}^2}{{\mathrm{x}}^3{\ln }^{3}3}=0$$

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