Prove that if lim f(x) = L1 and lim g(x) = L2, then lim (f(x))^(g(x)) = L1^L2 I am trying to prove that if
$$
\lim_{x \to c} (f(x)) = L_1
\\ \lim_{x \to c} (g(x)) = L_2
\\ L_1, L_2 \geq 0
$$
Then
$$
\lim_{x \to c} f(x)^{g(x)} = (L_1)^{L_2}
$$
I am doing this for fun, and my prof said that it shouldn't be too hard, but all I got so far is
$$
\forall \epsilon >0 \ \exists \delta > 0 : \text{if}\ \ |x-c|<\delta\ \ \ \text{then}\ |P(x)-L|<\epsilon
\\ |f(x)^{g(x)} - (L_1)^{L_2}| < \epsilon
$$
I have no idea how to proceed. Can someone help me out? I started by defining h(x) as $$(f(x))^{(g(x))}$$ but I couldn't go anywhere with that without basically defining the limit of h(x) as x approaches c to be L1^L2
 A: Given
$$
\lim_{x \to c} (f(x)) = L_1
\\ \lim_{x \to c} (g(x)) = L_2
$$
as long as $L_1 > 0$, then your statement is true as follows:
$$\lim_{x \rightarrow c} f(x)^{g(x)} = \lim_{x \rightarrow c} e^{\ln f(x)^{g(x)}} = \lim_{x \rightarrow c} e^{g(x) \ln f(x)}$$
Now we take limit of a composed function, and since the exponential function is continuous on $\mathbb{R}$,
$$\lim_{x \rightarrow c} e^{g(x) \ln f(x)} = e^{\lim_{x \rightarrow c}g(x) \ln f(x)}$$
Lets focus on the computing the exponent...
$$\lim_{x \rightarrow c}[g(x) \ln f(x)] = \lim_{x \rightarrow c}[g(x)] \cdot \lim_{x \rightarrow c} [\ln f(x)] = \lim_{x \rightarrow c}[g(x)] \cdot \ln \lim_{x \rightarrow c} [f(x)]$$
Note the expression $\ln \lim_{x \rightarrow c} [f(x)] = \ln L_1$ is the reason we must restrict $L_1 > 0$. We continue
$$\lim_{x \rightarrow c}[g(x)] \cdot \ln \lim_{x \rightarrow c} [f(x)] = L_2 \cdot \ln L_1 = \ln L_1^{L_2} $$
Now, lets take into the account the base with the exponent...
$$e^{\lim_{x \rightarrow c}g(x) \ln f(x)} = e^{\ln L_1^{L_2}} = L_1^{L_2}$$
Hence, 
$$\lim_{x \rightarrow c} f(x)^{g(x)} = L_1^{L_2}$$
A: It is often convenient to write $0^0=1,$ for example, in "Let $p(x)=\sum_{j=0}^n a_jx^j$  " it is assumed that $a_0x^0=a_0$ when $x=0.$
But if $L_1=L_2=0$ then $f(x)/g(x)$ can converge to any non-negative value,  or fail to converge.  Examples: Let $c=0:$ 
(1). Let $f_1(x)=1/e^{1/|x|}$ for $x\ne 0$ and $g_1(x)=|x|.$ Then $f_1(x)^{g_1(x)}=e$ for all $x\ne 0.$
(2). Let $f_2(x)=g_2(x)=|x|$ for $x\ne 0.$ Put $|x|=1/y.$ Then $y\to \infty$ as $x\to 0,$ and $f_2(x)^{g_2(x)}=1/(y^{1/y})=\exp ((\log y)/y).$ Now $(\log y)/y \to 0$ as $y\to \infty$ so $f_2(x)^{g_2(x)}\to 1$ as $x\to 0.$ 
(3). From examples (1) and (2), let $f_3(x)=f_1(x)$ when $1/x\in \mathbb Z$ and $f_3(x)=f_2(x)$ when $1/x \not \in \mathbb Z.$ Let $g_3(x)=g_1(x)=g_2(x).$ Then $f_3(x)^{g_3(x)}$ does not converge as $x\to 0$.
The  main result is valid for $L_1>0.$ Because   $\log f(x)^{g(x)}=g(x)\log f(x)$  whenever $|x-c|$ is small enough, and $\log f(x)$ will converge to $\log L_1.$  So $\log f(x)^{g(x)}$ will converge to $L_2\log L_1.$ 
