Functions of Several Variables — Question 9

Question 9 – Solution

Strategy Bound the absolute value by a constant times the distance to the origin.

Estimate For $(x,y)\ne(0,0)$, $$|f(x,y)|=|x|\frac{|x^2-y^2|}{x^2+y^2}\le |x|\le\sqrt{x^2+y^2}.$$ Therefore $|f(x,y)-0|\le r$, where $r$ is the distance to the origin.

Epsilon argument Given $\varepsilon>0$, choose $\delta=\varepsilon$. If $0<r<\delta$, then $|f(x,y)-f(0,0)|<\varepsilon$. Hence $\lim f=0=f(0,0)$ and $f$ is continuous.