I have been solving the limits of functions of two variables to review for my upcoming test.
I have two solutions with different answers. I am trying to review both solutions, but it seems both have no errors.
I tried to solve the
$\lim_{(x,y) \to (0,0)} \frac{x^2}{x - y}$
Here are my solutions.
Solution 1: Parameterization the equation $f(x,y) = \frac{x^2}{x-y}$ by $x(t)$ and $y(t)$
Let $k$ be a real number such that
$\frac{1}{k}=\frac{x^2}{x-y}$ Therefore, the said equation can be parameterized as $ \left\{ \begin{array}{lr} x(t)=t \\ y(t)=t-kt^2 \end{array} \right. $
This follows that
$\lim_{(x,y) \to (0,0)} \frac{x^2}{x - y}$
$ = \lim_{t \to 0} \frac{t^2}{t-(t-kt^2)}$
$ = \lim_{t \to 0} \frac{t^2}{kt^2}$
$=\frac{1}{k} = \left\{ \begin{array}{lr} 1&if&k=1 \\ -1&if&k=-1 \end{array} \right. $
Therefore, the limit does not exist.
Solution 2:
Parameterization the equation $f(x,y) = \frac{x^2}{x-y}$ by $x = r \cos(k)$ and $y = r \sin(k)$.
$\lim_{(x,y) \to (0,0)} \frac{x^2}{x - y}$
$ = \lim_{r \to 0} \frac{r \cos^2(k)}{[2\cos(k)+\sin(k)]}$
$=0$
Therefore, the limit is equal to 0.
My next question is when is the right time to parameterize the function in terms of $x = t$ and in terms of $x = r\cos(k)$ , $y = r\sin(k)$?
Please feel free to share your feedback about my two solutions.
In Solution 2 you cannot treat $k$ as a constant (independent of $r$). By doing so you are looking only at radial limits, i.e limits along lines approaching $(0,0)$.