# Thread: A limit

1. ## A limit

Assuming that :

$\lim_{x\to a}f(x) = l\neq 0$,and

$\lim_{x\to a}g(x)= m>0$ ,then

Prove that:

$\lim_{x \rightarrow a} f(x)^{g(x)} = \left( \lim_{x \rightarrow a} f(x) \right)^{\left(\lim_{x \rightarrow a} g(x) \right)}$

by using the ε-δ definition of a limit:

So we must prove that:

Given an ε>0 we must find a δ>0 such that:

for all ,x : 0<|x-a|<δ $\Longrightarrow | f(x)^{ g(x) } -l^m|<\epsilon$.

The point now is how do we manipulate , $|f(x)^{g(x)}-l^m|$

2. we may use
$f(x)^{g(x)}=e^{g(x)\ln f(x)}$

$|e^{g(x)\ln f(x)}-l^m|=|l^m||\frac{e^{g(x)\ln f(x)}}{l^m}-1|=$

$=|l^m||\frac{e^{(m+dp)\ln {(l+dq)}}}{l^m}-1|=$

$=|l^m||\frac{e^{(m+dp)\ln {l(1+ \frac{dq}{l})}}}{l^m}-1|=$

$=|l^m||\frac{ e^{ (m+dp)( \ln l+ \frac{dq}{l}) } }{ l^m }-1|=$

$=|l^m||\frac{ e^{ m \ln l +dp \ln l + \frac{mdq}{l} } }{ l^m }-1|=$

$=|l^m||\frac{ l^m e^{ { dp \ln l + \frac{mdq}{l}} } }{ l^m }-1|=$

$=|l^m|| { {dp \ln l} + \frac{mdq}{ l }} | <$

$< |l^m| ( |{dp \ln l}| +| \frac{mdq}{ l }|) < | dd | |l^m| (| \ln l | + |\frac{m}{l} |) < \varepsilon$

$| dd | < \frac{\varepsilon}{ |l^m|(| \ln l | + |\frac{m}{l} |) }$
where
$
dd=max(dp,dq)
$

3. Originally Posted by zzzoak
we may use
$f(x)^{g(x)}=e^{g(x)\ln f(x)}$

$|e^{g(x)\ln f(x)}-l^m|=|l^m||\frac{e^{g(x)\ln f(x)}}{l^m}-1|=$

$=|l^m||\frac{e^{(m+dp)\ln {(l+dq)}}}{l^m}-1|=$

$=|l^m||\frac{e^{(m+dp)\ln {l(1+ \frac{dq}{l})}}}{l^m}-1|=$

$=|l^m||\frac{ e^{ (m+dp)( \ln l+ \frac{dq}{l}) } }{ l^m }-1|=$

$=|l^m||\frac{ e^{ m \ln l +dp \ln l + \frac{mdq}{l} } }{ l^m }-1|=$

$=|l^m||\frac{ l^m e^{ { dp \ln l + \frac{mdq}{l}} } }{ l^m }-1|=$

$=|l^m|| { {dp \ln l} + \frac{mdq}{ l }} | <$

$< |l^m| ( |{dp \ln l}| +| \frac{mdq}{ l }|) < | dd | |l^m| (| \ln l | + |\frac{m}{l} |) < \varepsilon$

$| dd | < \frac{\varepsilon}{ |l^m|(| \ln l | + |\frac{m}{l} |) }$
where
$
dd=max(dp,dq)
$
Thanks ,but i am completely lost.

Where do you get dp,dq ??