Question Number 189870 by uchihayahia last updated on 23/Mar/23
$$ \\ $$$$\:{how}\:{do}\:{i}\:{prove}\:{this}\:{using}\:\epsilon\:{and}\:\delta \\ $$$$\:{please}\:{help} \\ $$$$\underset{{x},{y}\rightarrow\left(\mathrm{2},\mathrm{1}\right)} {\mathrm{lim}}\:{x}^{\mathrm{2}} {y}=\mathrm{4} \\ $$$$ \\ $$
Answered by mehdee42 last updated on 23/Mar/23
$${we}\:{have}\:{to}\:{show}\:: \\ $$$$\forall\epsilon>\mathrm{0}\:\exists\:\delta>\mathrm{0}\:\:;\:\parallel\left({x},{y}\right)−\left(\mathrm{2},\mathrm{1}\right)\parallel<\delta\Rightarrow\mid{x}^{\mathrm{2}} {y}−\mathrm{4}\mid<\epsilon \\ $$$${if}\:\delta_{\mathrm{1}} =\mathrm{1}\:\Rightarrow\:\mathrm{2}<{x}<\mathrm{3}\:,\:\mathrm{1}<{y}<\mathrm{2} \\ $$$$\mid{x}^{\mathrm{2}} {y}−\mathrm{4}\mid=\mid{xy}\left({x}−\mathrm{2}\right)+\mathrm{2}{x}\left({y}−\mathrm{1}\right)+\mathrm{2}\left({x}−\mathrm{2}\right)\mid−\leqslant\mid{xy}\mid\mid{x}−\mathrm{2}\mid+\mathrm{2}\mid{x}\mid\mid{y}−\mathrm{1}\mid+\mathrm{2}\mid{x}−\mathrm{2}\mid\leqslant\mathrm{8}\mid{x}−\mathrm{2}\mid+\mathrm{6}\mid{y}−\mathrm{1}\mid\leqslant \\ $$$$\mathrm{8}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{6}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{14}\parallel\left({x},{y}\right)−\left(\mathrm{2},\mathrm{1}\right)\parallel<\epsilon\Rightarrow\delta_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{14}}\epsilon \\ $$$$\Rightarrow\delta={min}\left\{\delta_{\mathrm{1}} ,\delta_{\mathrm{2}} \right\} \\ $$$$ \\ $$
Commented by uchihayahia last updated on 23/Mar/23
$${thank}\:{you},\:{why}\:\mid{xy}\mid\:{became}\:\mathrm{8}? \\ $$
Commented by mehdee42 last updated on 23/Mar/23
$$\mid{x}\mid<\mathrm{3\&}\mid{y}\mid<\mathrm{2}\Rightarrow\mid{x}\mid\mid{y}\mid\mid{x}−\mathrm{2}\mid<\mathrm{6}\mid{x}−\mathrm{2}\mid \\ $$$$\Rightarrow\mid{x}\mid{y}\mid\mid{x}−\mathrm{2}\mid+…+\mathrm{2}\mid{x}−\mathrm{2}\mid<\mathrm{6}\mid{x}−\mathrm{2}\mid+…+\mathrm{2}\mid{x}−\mathrm{2}\mid=\mathrm{8}\mid{x}−\mathrm{2}\mid+… \\ $$