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|$$<0.1$ .. whch will lead us to :
$0.9<e^x<1.1$
$ln0.9<x<ln1.1$
Then what to do ?? I am lost !
note : the answer should in terms of $ln$
 A: The function $f(x)=e^x$ is strictly increasing. 
At $x=0$, $f(x)$ definitely satisfies your property, since
$$|f(0)-1| = |e^0-1|= 0 <0.1$$
We'll look at two cases: when $x$ is positive, and when $x$ is negative.

Case 1: $x$ is positive.
When $x>0$, $e^x > 1$, so $|f(x)-1| =f(x)-1$. Now what is the maximum positive $x$, call it $x_{\text{max}}$, such that $f(x)-1 \leq 0.1$? This is simply the solution to $$e^x-1=0.1$$
Using $\ln$ for the natural logarithm, this gives
$$x_{\text{max}} = \ln 1.1$$
Case 2: $x$ is negative.
When $x<0$, $e^x<1$, so $|f(x)-1| = 1-f(x)$. Now what is the minimum negative $x$, call it $x_{\text{min}}$, such that $1-f(x) \leq 0.1$? This is the solution to
$$e^x=0.9$$
This gives
$$x_{\text{min}}=\ln 0.9$$

Now we've found the minimum negative value of $x$ for which your equation works, and the maximum positive value. The maximum $\delta$ for which $|x|<\delta$ implies $|f(x)-1|<0.1$ will be the minimum of $-x_{\text{min}}$ and $x_{\text{max}}$. $$\delta=\min\{-\ln 0.9, \ln 1.1\}$$
The natural logarithm has the nice property that for all $x>0$, $- \ln x = \ln (1/x)$. This means that $- \ln 0.9 = \ln 1.\overline{1}$. The natural logarithm is also strictly increasing where it's defined (on the positive real numbers), so $\ln 1.\overline{1} > \ln 1.1$. This means $$\delta = \ln{1.1} \approx 0.09531$$
A: 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 $|x|\ge\ln(11/10)$?
A: 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, $\delta=\ln(1.1)$. 
