<s>[INST] Continuity of function of two variable by Epsilon-Delta technique. Given a function
$f(x,y)= \frac{x^3y^2+x}{x^2+y^2}$ when $(x,y) \neq (0,0)$. Show that the function $f(x,y)$ is continuous at $(1,1)$ by Epsilon-Delta definition.
I can show it easily by the fraction of two continuous functions is continuous. But I need to prove it directly by Epsilon-Delta technique. I'm not able to think, how to do it. Please give some hint!
Effort: For any given $\epsilon>0$, we have
$|f(x,y)-1|= |\frac{x^3y^2+x}{x^2+y^2}-1|=|\frac{x^3y^2+x-x^2-y^2}{x^2+y^2}|$
Now, what's next?
 [/INST] For any $\epsilon > 0$ there exists a $\delta>0$ such that
$\sqrt {(x-1)^2 + (y-1)^2} < \delta \implies \left | \frac {x^3y^2 + x - x^2 - y^2}{x^2+y^2}\right | <\epsilon$
The numerator can be simplified and factored.
$(x^3-1)y^2 - x(x-1) = (x-1)((x^2+x+1)(y^2)-x)$
$(x-1) < \delta$
Let $\delta\le 1$
$|x|< 2, |y| < 2$
$|((x^2+x+1)(y^2)-x)| \le (4 + 2 + 1)(4) + 2 \le 30$
$x^2+y^2\ge 1$
$\left | \frac {((x^2+x+1)y^2-x)(x-1)}{x^2+y^2}\right | < 30\delta \le \epsilon$
$\delta = \min (1,\frac {1}{30}\epsilon)$
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