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
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 $L_2\log L_1.$