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 $ln$

You are almost done.
Your last inequality is
$$\ln\frac9{10}<x<\ln\frac{11}{10}$$
or
$$-\ln\frac{10}9<x<\ln\frac{11}{10}$$
Since $11/10<10/9$, for $|x|<\delta=\ln(11/10)$, the inequality $|f(x)-1|<0.1$ holds. What happens if $|x|\ge\ln(11/10)$?
