Limit Rule $\lim f(x)^{g(x)}$

Assuming $f(x)>0$ for some open set surrounding $x_0$, we know that because the natural logarithm is continuous on $(0,\infty)$ that \begin{align} \ln \left( \lim_{x\rightarrow x_0}{f(x)^{g(x)}}\right) & =\lim_{x\rightarrow x_0}{\ln \left(f(x)^{g(x)}\right)} \\ & =\lim_{x\rightarrow x_0}{g(x)\ln f(x)} \\ & =\left(\lim_{x\rightarrow x_0}{g(x)} \right)\left(\lim_{x\rightarrow x_0}{\ln f(x)}\right) \\ & =\left(\lim_{x\rightarrow x_0}{g(x)}\right)\ln \left( \lim_{x\rightarrow x_0}{f(x)}\right) \\ & =\ln \left[\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}\right] \end{align}

Then, because the natural logarithm is injective, we find that $$\lim_{x\rightarrow x_0}{f(x)^{g(x)}}=\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}$$

Yes. Note that the function $f(x,y)=x^y=e^{y\ln(x)}$ is continuous for $x,y>0$, so if $(x_n,y_n)\to (x,y)$ with $x,y>0$ then $f(x_n,y_n)\to f(x,y)$.

Because

1. $y(x)=ln(x)$ is continuous at $x = c > 0$

2. $\lim\limits_{x\to 0}f(x) = c$

according to the composition law, we have

$$\lim\limits_{x \to 0}lnf(x) = ln\lim\limits_{x \to 0}f(x) = lnc$$

Because $\lim\limits_{x \to 0}g(x) = d$, we have

$$\lim\limits_{x\to 0}g(x)lnf(x) = \lim\limits_{x\to 0}g(x)\cdot\lim\limits_{x \to 0}lnf(x) = dlnc$$

Apply composition law again, we get

$$\lim\limits_{x\to 0}f(x)^{g(x)} = \lim\limits_{x\to 0}e^{g(x)lnf(x)} = e^{\lim\limits_{x\to 0}g(x)lnf(x)} = e^{dlnc} = c^d$$