Chapter 14: Partial Derivatives - Section 14.2 - Limits and Continuity in Higher Dimensions - Exercises 14.2 - Page 798: 70

$|f(x,y) -f(0,0) | \lt \epsilon $

We have $f(x,y)=\dfrac{y}{x^2+1}$ and $f(0,0)=0$ Now, $|f(x,y) -f(0,0) | \lt \epsilon $ or, $|\dfrac{y}{x^2+1}-0| \lt 0.05 $ This implies that $\dfrac{|y|}{x^2+1} \lt 0.05$ Since,$\dfrac{|y|}{x^2+1} \leq \sqrt {x^2+y^2 }$ Now, $\sqrt {x^2+y^2 } \lt 0.05$ Suppose $\delta =0.05$ So , $\sqrt {x^2+y^2 } \lt \delta$ Thus, $|f(x,y) -f(0,0) | \lt \epsilon $