Question Number 69075 by Aditya789 last updated on 19/Sep/19
Commented by Prithwish sen last updated on 19/Sep/19
$$\mathrm{Divide}\:\mathrm{both}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator} \\ $$$$\mathrm{by}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 19/Sep/19
$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{5}\right)\:={x}^{\mathrm{3}} \:+{p}\left({x}\right)\:{with}\:{deg}\left({p}\right)\leqslant\mathrm{2} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{7}\right)\:={x}^{\mathrm{3}} \:+{q}\left({x}\right)\:{with}\:{deg}\left({a}\right)\leqslant\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\overset{−} {+}\infty} \frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{5}\right)}{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{7}\right)}\:={lim}_{{x}\rightarrow\overset{−} {+}\infty} \:\:\:\:\frac{{x}^{\mathrm{3}} \:+{p}\left({x}\right)}{{x}^{\mathrm{3}} \:+{q}\left({x}\right)} \\ $$$$={lim}_{{x}\rightarrow\overset{−} {+}\infty} \:\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\:=\mathrm{1}\:. \\ $$