Question Number 120194 by huotpat last updated on 30/Oct/20
Answered by MJS_new last updated on 30/Oct/20
$$\mathrm{let}\:{y}={px}\wedge{z}={qx}\:\mathrm{with}\:{p},\:{q}\:\in\mathbb{R} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{pqx}^{\mathrm{3}} }{\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} }\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{pqx}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{1}}\:=\mathrm{0} \\ $$