evaluating limits with 2 variables

Question

Evaluate the limit :

$\lim_{(x,y)\to (2,2)}$ $\frac{x^2 + y^2 - 8}{\sqrt{x^2 +y^2} - \sqrt{8}}$

1. $−1$

2. $\infty$

3. $0$

4. $1$

5. none of the other choices

6. does not exist

How is the answer number 5, I thought it should be "does not exist"?

Answer 1

hint

$$x^2+y^2-8=$$ $$(\sqrt{x^2+y^2}+\sqrt{8})(\sqrt{x^2+y^2}-\sqrt{8})$$

You will find $$2\sqrt{8}.$$

Answer 2

HINT: If permissible, you may convert the expression to polar coordinates. We know that $x = r\cos\theta$, $y = r\sin\theta$, and hence $r^2 = x^2 + y^2$. So, $$\require{cancel}{x^2 + y^2 - 8 \over \sqrt{x^2 + y^2} - \sqrt{8}} = {r^2 - 8\over r - \sqrt8} = {\cancel{(r-\sqrt{8})}(r + \sqrt{8})\over \cancel{r-\sqrt8}} = r+\sqrt{8}.$$