I need to prove the continuity of $f(x)=\log x$ using a $\epsilon-\delta$ proof I need to prove the  continuity of $f(x)=\log x$ using a $\epsilon-\delta$ proof
These is what I have so far but am not sure how to continue
$|\log x-\log a| < \epsilon$ 
$\log a- \epsilon < \log x < \log a+ \epsilon$
$\frac{a}{e^\epsilon} < x < {a}e^\epsilon$
Any help is appreciated 
 A: By your inequality, the absolute value of the difference is $\lt \epsilon$ if 
$$\frac{a}{e^{\epsilon}}-a \lt x-a\lt ae^\epsilon -a$$
(we subtracted $a$ from each side of each of your two inequalities).
Let $\delta=a\min\left(1-\frac{1}{e^{\epsilon}}, e^\epsilon -1\right)$. 
Remark: Actually, $1-\frac{1}{e^{\epsilon}}$ is the smaller of the two, so in effect we are letting that be $\delta$. But we really don't need to bother finding that out: all we need to do is to show there is a $\delta$ that works.  
A: Hint: from the right (i.e., $\,x>a\,$):
$$|\log x-\log a|=\log\frac{x}{a}<\epsilon\Longleftrightarrow \frac{x}{a}<e^\epsilon\Longleftrightarrow x<ae^\epsilon\;\;(\text{remember}:\;a,x>0\,\;!)\Longrightarrow$$
$$x-a<a(e^\epsilon -1)$$
and there you have your $\,\delta>0\ldots\,$
A: Show it at $x = 1$. Then spread it around using the fact that $\log(xy) = \log(x) + \log(y)$.
A: What you are trying to prove is that for any fixed $x$, $\forall\, \epsilon>0\;\; \exists\, \delta>0$ such that $|\log(x+\delta)-\log x|<\epsilon$.
So, you need to find the $\delta$ in terms of $\epsilon$ and $x$.
