Let $\lim_{k \to \infty} x_k = a$ and $\lim_{k \to \infty} y_k = b$. Does $\lim_{k \to \infty} {x_k}^{y_k} =a^b$ hold? Assume that $(x_k)$ is real or complex sequence, and $(y_k)$ is real sequence such that $\lim_{k \to \infty} x_k = a$ and $\lim_{k \to \infty} y_k = b$.


*

*I would like to ask if $\lim_{k \to \infty} {x_k}^{y_k} =a^b$ holds.

*If not, does it hold in case $b=1$?
Thank you for your help!
 A: Grey Fox's answer is right about the real case. Specifically, when $x_k > 0$, $a > 0$, and $y \in \mathbb{R}$, we have $\lim_{k \to \infty} {x_k}^{y_k} = a^b$. 
However, if $a \leq 0$, $\lim_{k \to \infty}{x_k}^{y_k}  = a^b$ may fail to be true or even to make sense. For a simple example, take $x_k = \frac{1}{k}$ $y_k = 0$. Then $x_k \to 0$, $y_k \to 0$, and ${x_k}^{y_k} = \left(\frac{1}{k}\right)^0 = 0 \to 0$, but $0^0$ is not defined. 
If the $x_k$ are not real, then ${x_k}^{y_k}$ will not even be single-valued, so, unless we agree on a branch, it doesn't make sense to talk about $\lim_{k \to \infty} {x_k}^{y_k}$. If you wanted, I suppose you could talk about the limit points of the set $\{{x_k}^{y_k}\}$, but the sequence won't have a limit in the traditional sense. 
If we require $a > 0$ and $x_k$ real, then we will have $\lim_{k \to \infty} {x_k}^{y_k} = a^b$ (ignoring finitely many terms that may not be defined). To see this, note that $a > 0$ implies that, for all but finitely many $k$, $x_k > 0$. We compute
$$
{x_k}^{y_k} = e^{y_k \log x_k} = e^{y_k (\log |x_k| + i \arg x_k)} = e^{y_k (\log |x_k| + i 0)} = e^{y_k \log|x_k|}
$$
This last expression is single-valued and continuous, hence
$$
\lim_{k \to \infty} {x_k}^{y_k} = e^{(\lim_{k \to \infty} y_k) \log |\lim_{k \to \infty} x_k|} = e^{b \log |a|} = e^{b \log a} = a^b
$$
A: Take $f(x,y)=x^y$. Then $g(x,y)=\ln f(x,y)=y \ln x$ is continuous. The continuity of $h(x)=e^x$ implies that $(h\circ g)(x,y)= f(x,y)$ is continuous. Therefore taking the limit $\lim f(x_n,y_n)=f(a,b)=a^b$.
