Question Number 28097 by tawa tawa last updated on 20/Jan/18
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}^{\mathrm{x}} \:−\:\mathrm{2}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by abdo imad last updated on 20/Jan/18
$$={lim}_{{x}\rightarrow+\propto} \:\:\frac{\mathrm{3}^{{x}} \left(\mathrm{1}\:−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \right)}{{x}^{\mathrm{2}} }={lim}_{{x}\rightarrow+\propto} \:\:\frac{\mathrm{3}^{{x}} }{{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow+\propto\:\:} {e}^{{xln}\mathrm{3}} \:{e}^{−\mathrm{2}{lnx}} =\:{lim}_{{x}\rightarrow+\propto} \:\:{e}^{{x}\left({ln}\mathrm{3}\:−\mathrm{2}\frac{{lnx}}{{x}}\right)} \\ $$$$={lim}_{{x}\rightarrow+\propto} \:\:{e}^{{xln}\mathrm{3}} =+\propto\:\:\:{because}\:\:\:{lim}_{{x}\rightarrow\propto} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{0} \\ $$$${and}\:{lim}_{{x}\rightarrow+\propto} \:\:\frac{{lnx}}{{x}}=\mathrm{0}\:. \\ $$
Commented by çhëý böý last updated on 20/Jan/18
$${Good}\:{work}\:{sir} \\ $$$$ \\ $$
Commented by tawa tawa last updated on 21/Jan/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$