TITLE: Limit Rule $\lim f(x)^{g(x)}$
QUESTION [8 upvotes]: 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 ?

REPLY [2 votes]: 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
$$\lim\limits_{x\to 0}f(x)^{g(x)} = \lim\limits_{x\to 0}e^{g(x)lnf(x)} = e^{\lim\limits_{x\to 0}g(x)lnf(x)} = e^{dlnc} = c^d$$