Question Number 188985 by normans last updated on 10/Mar/23
$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{Lim}}_{\boldsymbol{{x}}\rightarrow\sim} \:\:\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{4}}{\mathrm{9}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}}\:=\:??\: \\ $$
Commented by MJS_new last updated on 10/Mar/23
$$\underset{{x}\rightarrow\pm\infty} {\mathrm{lim}}\:\frac{{ax}^{{n}} +{c}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +…+{c}_{{n}} {x}^{\mathrm{0}} }{{bx}^{{n}} +{d}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +…{d}_{{n}} {x}^{\mathrm{0}} }\:=\frac{{a}}{{b}} \\ $$
Answered by cortano12 last updated on 10/Mar/23
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{4x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{4}}{\mathrm{9x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{9}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{3}} \left(\mathrm{4}−\frac{\mathrm{2}}{\mathrm{x}}−\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{3}} }\right)}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{9}−\frac{\mathrm{4}}{\mathrm{x}}+\frac{\mathrm{9}}{\mathrm{x}^{\mathrm{3}} }\right)} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$