Vtyjkyuk

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:
$|x^2+y^2| < 0.01$
$sqrtx^2+y^2 < sqrt0.01$
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, $|x^2+y^2-1| < 0.01$
$sqrtx^2+y^2 < sqrt0.01$.

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

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

If $|x^2+y^2-1|<0.01$, you cannot conclude that $sqrtx^2+y^2 < sqrt0.01$. For example, taking $x=1, y=0$, you can see that $$|x^2+y^2-1| =0<0.01$$ clearly holds, however, $$sqrtx^2+y^2 = 1 < sqrt0.01$$ does not hold.

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

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 $0$