## Thomas' Calculus 13th Edition

$|f(x,y) -f(0,0) | \lt \epsilon$
We have $f(x,y)=\dfrac{x+y}{x^2+1}$ and $f(0,0)=0$ Now, $|f(x,y) -f(0,0) | \lt \epsilon \implies |\dfrac{x+y}{x^2+1}-0| \lt 0.01$ This implies that $\dfrac{|x+y|}{x^2+1} \lt 0.01$ Since,$|x+y| \leq 2 \sqrt {x^2+y^2 }; x^2+1 \geq 1$ So, $\dfrac{|x+y|}{x^2+1} \leq 2 \sqrt {x^2+y^2}$ This implies that $2 \sqrt {x^2+y^2} \lt 0.01 \implies \sqrt {x^2+y^2} \lt 0.005$ Suppose $\delta =0.005 \implies \sqrt {x^2+y^2 } \lt \delta$ Thus, $|f(x,y) -f(0,0) | \lt \epsilon$