Proving a function(with two variables) is continuous I am having a difficulty solving this problem. First of all, I'm sorry that the problem isn't well written but I am not very good with typing out math problems, due to the fact that I am new to this, so I hope it's at least understandable. Next, I want to say that I've tried solving this using polar coordinates and also by changing y with  $y=kx$, $y=kx^2$. But it didn't work. Anyway, the problem says: Find parameter a so that a function is continuous. (I tried to translate it correctly in English.) I hope someone can help me solve this problem, I would be so grateful.
$$
f(x,y) =
 \begin{cases} 
      \dfrac{5 - \sqrt{25-x^2-y^2}}{7 - \sqrt{49-x^2-y^2}} & (x,y)\neq (0,0) \\
        \\
      a & (x,y)=(0,0) 
   \end{cases}
$$
 A: In these problems with roots, a typical strategy is the one of “rationalize” the fraction:
If you write:
$$\frac{5-\sqrt{25-x^2-y^2}}{7-\sqrt{49-x^2-y^2}}\cdot \frac{5+\sqrt{25-x^2-y^2}}{5+\sqrt{25-x^2-y^2}}\cdot \frac{7+\sqrt{49-x^2-y^2}}{7+\sqrt{49-x^2-y^2}}= \frac{x^2+y^2}{x^2+y^2}\cdot\frac {7+\sqrt{49-x^2-y^2}} {5+\sqrt{25-x^2-y^2}}=\frac {7+\sqrt{49-x^2-y^2}} {5+\sqrt{25-x^2-y^2}}$$
So your limit is $\frac 75$!
A: Hint:
You only need to determine, in polar coordinates,  $\;\displaystyle\lim_{r\to 0}\,\dfrac{5 - \sqrt{25-x^2-y^2}}{7 - \sqrt{49-x^2-y^2}}$.
Rewrite this fraction, with the conjugate expressions, as
$$\frac{5 - \sqrt{25-r^2}}{7 - \sqrt{49-r^2}}=\frac{5^2 - (25-r^2)}{5 + \sqrt{25-r^2}}\, \frac{7 +\sqrt{49-r^2}}{7^2 -(49-r^2)}.$$
Can you end the calculations?
A: Set $r^2=x^2+y^2$;
$5(1-r^2/25)^{1/2}=$
$5(1-(1/2)(r^2/(25))+ O(r^4))$;
$7(1-r^2/49)^{1/2}=$
$7(1-(1/2)(r^2/(49))+O(r^4))$;
Numerator: $r^2/(10)+O(r^4)$;
Denominator: $r^2/(14)+O(r^4)$;
Take the limit $r^2 \rightarrow \infty.$
$a=?$
