Show whether or not the functions $f(x,y)=|x|^y$, $g(x,y)=|x|^{|1/y|}$ have limits at $(0,0)$.

Question

Show whether or not the functions $f(x,y)=|x|^y$, $g(x,y)=|x|^{|1/y|}$ have limits at $(0,0)$.

By the answers only the latter has a limit at $(0,0)$ but I don't know how to prove.

Answer 1

I might as well write an answer.

For the first question, we note that if $\lim_{(x,y) \to (0,0)} f(x,y)$ were to exist and equal $L$, then for every pair of subsequences $x_n \to 0,y_n \to 0$ we should have $f(x_n,y_n) \to L$ (as a sequence). However, considering for example $(x_n,y_n) = (0,\frac 1n)$ we have $f(x_n,y_n) = 0^{\frac 1n} = 0 \to 0$ ,and with $(x_n,y_n) = (\frac 1n,0)$ we have $f(x_n,y_n) = \left(\frac{1}{n}\right)^0 = 1 \to 1$, and $0 \neq 1$ so no candidate for $L$ can exist.

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For the second question, we have tho show that the limit of $|x|^{\frac 1{|y|}}$ exists at $(0,0)$. By travelling along the $x$ axis, a guess of the limit value is $0$.

To show that the limit is zero, fix $\epsilon > 0$. Suppose that $|x| < \min\{\frac 12,\epsilon\}$(nothing special about half, can take anything smaller than one) and $|y| < 1$. Then, $|x| < 1$, so if $a>b$ then $|x|^a < |x|^b$. Furthermore, $|y| < 1$ means $\frac 1{|y|} > 1$, which means that $|x|^{\frac 1{|y|}} < |x| < \epsilon$.

We have shown therefore, that $|x| < \min\{\frac 12,\epsilon\}$ and $|y|<1$ implies $|x|^{\frac 1{|y|}}<\epsilon$. Can you do it from here?