How to calculate limit? (Two variables)

Question

I can't find the way to calculate this limit. Is it so difficult? Thanks for helping

$\lim_{(x,y)\rightarrow(0,0)} \frac {x^2}{x^2-y}$

Answer 1

You want to convert the limit you have to polar, so take the limit as $r$ approaches $0$. Convert using $x=rcos(\theta)$ and $y=rsin(\theta)$ and it should be a simple limit to evaluate in polar.

Answer 2

HINT

Let consider

• $x=0\implies \frac{x^2}{x^2-y}=?$

• $x=t \quad y=t^2-t^3\quad t\to 0^+ \implies \frac{x^2}{x^2-y}=\frac{t^2}{t^2-t^2+t^3}=\frac1t\to?$

Answer 3

Well, Im'm kinda new in this topic, but let me try; First of all there exists some methods, the first method it's a reference to one variable calculus "unilateral limits" and I'm going to use that to show that your limit does not exits:

Consider one coordinate axis, for example let x the first axis to set: $ \lim_{(x,y)\rightarrow (0,0) }f(x,y)$

Then: $\lim_{(x\rightarrow 0)} (\lim_{y\rightarrow 0}(x^2/x^2-y))=L_1 $

$\lim_{(y\rightarrow 0)} (\lim_{x\rightarrow 0}(x^2/x^2-y))=L_2 $

You should know how to resolve those limits, but let me be more explicit: For the first limit, as long as y tends to 0 then: $\lim_{(x\rightarrow 0)} (x^2/x^2))=L_1=1$

For the other limit you should make the same proccess:. As long as x tends to 0 the limit changes in to another expresion $\lim_{(y\rightarrow 0)} (0/0-y))=L_2=0 $

From here should notice that $L_1\neq L_2$ so you could conclude that the limit as (x,y) tends to (0,0) of f(x,y) does not exist.

Another way to make this problem its by definition, but if you notice this by this method you should be able to say that the limit does not exists. I'm so sorry if my english it's not good enought to explain this.