Question Number 412 by 123456 last updated on 25/Jan/15
$$\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\:\frac{\mathrm{sin}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} \\ $$
Answered by prakash jain last updated on 31/Dec/14
$${f}\left({x},{y}\right)=\frac{\mathrm{sin}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{sin}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{case}\:\mathrm{0}<\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }<\delta \\ $$$$\mid{f}\left({x},{y}\right)−\mathrm{0}\mid=\mid\frac{\mathrm{sin}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\mid \\ $$$$=\mid\frac{\mathrm{sin}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\mid\:\mid\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\mid<\mid\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\mid<\delta \\ $$$$\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0} \\ $$