Proving a function(with two variables) is continuous

Question

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.

$$ f(x,y) = \begin{cases} \dfrac{5 - \sqrt{25-x^2-y^2}}{7 - \sqrt{49-x^2-y^2}} & (x,y)\neq (0,0) \\ \\ a & (x,y)=(0,0) \end{cases} $$

Answer 1

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 $\frac 75$!

Answer 2

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?

Answer 3

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 $r^2 \rightarrow \infty.$

$a=?$