# Giving an epsilon-delta proof to show that a function is continuous

I want to show that $$f(x,y)=\frac{y}{(x^2+y^2)}$$ is continuous on $$[0,1]\times[1,2]$$.

Let $$\epsilon >0$$ be given. I want to choose a $$\delta>0$$ such that for $$||(x_1,y_1)-(x_2,y_2)||<\delta$$ we have $$|\frac{y_1}{(x_1^2+y_1^2)}-\frac{y_2}{(x_2^2+y_2^2)}|<\epsilon$$.

The only thing I found is that $$\begin{equation}|\frac{y_1}{(x_1^2+y_1^2)}-\frac{y_2}{(x_2^2+y_2^2)}|\leqslant \frac{y_1(x_2^2+y_2^2)-y_2(x_1^2+y_1^2)}{(x_1^2+y_1^2)(x_2^2+y_2^2)}\leqslant \frac{2(1+4)-1(0+1)}{(0+1)(0+1)}=9.\end{equation}$$

I have no idea how to find such a $$\delta$$.

• It's not continuous ! $\lim_{t\to 0^+}f(0,t)=+\infty$. – Surb Oct 5 '18 at 9:06
• Correct with this domain, but that was a typo... I changed it now! – David W. Oct 5 '18 at 9:11

## 1 Answer

Your function is not defined at $$(0,0)$$. For the other points: fix $$a=(x_0,y_0)$$ and let $$\epsilon >0$$. Observe that if $$\delta < \|a\|/2$$ then for all $$b=(x_1,y_1)$$ such that $$\|a-b\|< \delta$$ we have $$\frac{1}{(\|a\|-\frac{\|a\|}{2})^2 }=\frac{4}{\|a\|^2} \geq \frac{1}{\|b\|^2}.$$ Observe that $$|y_0 - y_1|\leq \| a-b\|$$ and $$|f(a)-f(b)| = \left| \frac{y_0}{\|a\|^2} - \frac{y_1}{\|b\|^2} \right| = \left| \frac{ y_0 \|b\|^2 - y_1 \| a\|^2}{\|a\|^2 \|b\|^2}\right| \leq \left| \frac{ |y_0|\cdot \left|\|b\|^2 - \| a\|^2 \right|+ |y_1 - y_0|\cdot \|a\|^2}{\|a\|^2 \|b\|^2}\right| \leq \frac{4}{\|a\|^4}\left( |y_0|\cdot \left|\|b\|^2 - \| a\|^2 \right|+ |y_1 - y_0|\cdot \|a\|^2 \right).$$ Choose $$\delta < \min \left\{ \frac{2 \epsilon}{5\|a\|}, \frac{\|a\|}{2}, \frac{\epsilon}{\|a\|^2} \right\}.$$

• When I try to do this, I get $|f(a)-f(b)|\leqslant \frac{4|4y_0-y_1|}{||a||^2}$. Also, I don't get why you did say that $\delta <||a||/2$.. – David W. Oct 5 '18 at 10:51
• You are right. I fixed the proof. I choose $\delta < \|a\|/2$ simply to avoid (0,0) and make estimates easier in some parts. – Nick Bottom Oct 5 '18 at 12:02