If we have $f(x)=e^x$, then what is the maximum value of $δ$ such that $|f(x)-1|< 0.1$ whenever $|x|<δ$? If we have $f(x)=e^x$, then what is the maximum value of $δ$ such that $|f(x)-1|< 0.1$ whenever $|x|<δ$?
I tried to solve this problem with delta-epsilon definition
from the definition, 1 is L (the value of the limit)
so the limit will be :
$lim_{x--->0}$ $e^x=1$ ( the approching value will then be $0$ )
then,  $|e^x-1|$$<0.1$ .. whch will lead us to :
$0.9<e^x<1.1$
$ln0.9<x<ln1.1$
Then what to do ?? I am lost !
note : the answer should in terms of $ln$

Your deduction is perfect. From your derivation you get that $x<\ln 1.1$ and $\ln 0.9<x$, so we need to have $|x|<\min(\ln (1.1),|\ln(0.9)|)$. So, it remains to find $\delta=\min(\ln (1.1),|\ln(0.9)|)=\min(\ln(1.1),-\ln(0.9))$. 
But since $\ln(x)$ is concave, we have 
$$\frac{1}{2}(\ln(1-x)+\ln(1+x))<\ln(\frac{1}{2}(1+x+1-x))=\frac{1}{2}\ln(1)=0.$$ 
Or $\ln(1+x)<-\ln(1-x)$. Letting $x=0.1$, we get $\ln(1.1)<-\ln(0.9)$ and hence, $\delta=\ln(1.1)$. 
