So I've got the function $$f(x,y)=x^y$$ and need to prove that it does not have a limit.
I was given the hint that I should find a curve $y=\gamma(x)$ along which $f(x,y)$ is a constant value. I'm meant use this hint to show that along the curve $y=\gamma_K(x)$, $$\lim_{(x,y)\to(0,0)}f(x,y)=K$$ for any given positive $K$ value.
I really don't get what the hint is asking for; it actually makes the question even more confusing.
On
Hint:
Suppose $x^y = K$, hence $y\ln(x) = ln(K)$ so your curve should be something like $y=ln(K)/ln(x)$ but, this is not always well defined.
On
The hint means that you can approach $0$ from different paths (the curve) by choosing different sequences along which $f$ is constant to an arbitrary value $K$ and hence the limit of $f$ is also equal to $K$.
Taking that idea into action, can look like the following: Assume that the limit exists. Let $K>0$ be arbitrary. If you define a curve by $\gamma_K(x)=\frac{\log(K)}{\log(x)}$, then $f(x,\gamma_K(x))=K$ and thus $\lim_{x\searrow0}f(x,\gamma_K(x))=K$. Furthermore, $\log_{x\searrow0}x=0$ and $\lim_{x\searrow0}\gamma_K(x)=0$ and therefore $(x,\gamma_K(x))$ is a sequence converging to $0$. As $K$ is arbitary and in order for the limit to exist every sequence converging to $0$ must result in the same limit, this contradicts the existence of a limit.
In fact, you can use also two different sequences $a_n =(x_n,y_n)$ and $b_n=(u_n,v_n)$ which both tend to $(0,0)$ such that $\lim f(a_n)$ is different of $\lim f(b_n)$. I found it more easy to deal with than finding a "curve".
For instance, with $(x_n,y_n):=(\frac1n,\frac1n)$ and $(u_n,v_n):=(e^{-n},\frac1n)$, you will have $$ f(a_n) = \exp(y_n\ln(x_n))= \exp(-\tfrac{\ln(n)}{n}) \rightarrow \exp(0) =1 $$ and $$ f(b_n) = \exp(v_n\ln(u_n))= \exp(-\tfrac{n}{n}) \rightarrow \exp(-1) \neq 1 $$
Using a sequence instead a curve, shows maybe more easily that the point is a different "speed" of the different components.