Is the following reasoning sound for demonstrating that the limit does not exist. The limit is:
$$\lim_{(x,y) \to (0,0)}{x^y}$$
I show that the restriction gives me two different values of the limit as follows:
$$\lim_{(0,y) \to (0,0)}{x^y}=\lim_{(0,y) \to (0,0)}0^y=0$$
$$\lim_{(x,0) \to (0,0)}{x^y}=\lim_{(x,0) \to (0,0)}x^0=1$$
Therefore the limit does not exist.
Best Answer
It is necessary to choose a proper domain as the function $f(x, y) = x^y$ cannot be defined on all of $\mathbb{R}^2$. One of the most straightforward fixes is to take the domain to be $[0, \infty)\times [0, \infty)\setminus \{(0, 0)\}$.
Now suppose that $\lim_{(x, y)\rightarrow (0, 0)}f(x, y)$ exists and is equal to $l$. This implies that there exists a $\delta>0$ such that $|f(x, y)-l|<1/4$ whenever $||(x, y)||<\delta$. Choose $y$ such that $\delta>y>0$. For such a $y$, $|f(0, y)-l| = |l|<1/4$. Now choose $x$ such that $\delta >x>0$. For such an $x$, $|f(x, 0)-l| = |1-l|<1/4$. This gives us $|l|+|1-l|<1/2$, which using triangle inequality implies that $1<1/2$, a contradiction.