Limit Rule $\lim f(x)^{g(x)}$ I want to know the following is true : If $$ c,\ d\in {\bf R},\  \lim_{x\rightarrow 0} f(x)=c>0,\ \lim_{x\rightarrow 0} g(x) =d>0$$ then $$ \lim_{x\rightarrow 0} f(x)^{g(x)} = c^d$$
In calculus book such formula cannot be found. 
Consider the problem : $$\lim_{x\rightarrow 0} (1+\sin\ 4x)^{\cot\ x} $$
To find a limit, we must use ${\rm log}$ and L'Hospital. But some student suggests that $$ \lim_{x\rightarrow 0} (1+\sin\ 4x)^{\frac{1}{\sin\ 4x} \frac{\sin\ 4x}{\sin\ x}\cos\ x}=e^4$$
This argument is clear. But I know that it is informal. Is there a minus point in such way ?
 A: 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 $$\lim_{x\rightarrow x_0}{f(x)^{g(x)}}=\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}$$
A: 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 $f(x_n,y_n)\to f(x,y)$.
A: Because 


*

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

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