I need a justification for epsilon delta definition of limit

Question

The limiting value of $f(x,y)=x^2+y^2$ is $0$ as $(x,y)$ tends to $(0,0)$.

So if I want to prove the same using $\epsilon$ - $\delta$ definition for a given $\epsilon=0.01$, I will certainly go as follows: $|x^2+y^2| < 0.01$ $\sqrt{x^2+y^2} < \sqrt{0.01}$ so I got a value of $\delta=0.1$ corresponding to $\epsilon=0.01$.

Now my question is if lets say the limiting value of $f(x,y)=x^2+y^2$ is $1$ as $(x,y)$ tends to $(0,0)$ then, $|x^2+y^2-1| < 0.01$ $\sqrt{x^2+y^2} < \sqrt{0.01}$.

Here also exists a $\delta$ corresponding to $\epsilon$.

Does that mean $1$ is the limiting value of $f(x,y)$?

Answer 1

No. Observe that your inequality $|x^2+y^2-1| < 0.01 \sqrt{x^2+y^2}$ is false !

Answer 2

If $|x^2+y^2-1|<0.01$, you cannot conclude that $\sqrt{x^2+y^2} < \sqrt{0.01}$. For example, taking $x=1, y=0$, you can see that $$|x^2+y^2-1| =0<0.01$$ clearly holds, however, $$\sqrt{x^2+y^2} = 1 < \sqrt{0.01}$$ does not hold.

Answer 3

As you know limits are unique, so once you proved that the limit is $0$, then you can not prove that the limit is also $1$

In your case you have proved that $x^2+y^2 \to 0$ as $(x,y)\to (0,0)$, thus you can not prove the same limit is $1$ or anything other than $0$