$\varepsilon-\delta$ Proof: $\lim\limits_{x\to0}e^x=1$ I'm trying to learn how to do $\varepsilon-\delta$ proofs and am looking for verification that this proof that $\lim\limits_{x\to0}e^x=1$ is correct, as well as any ways that it could be improved. (Note: $f\left(x\right)=e^x$.)
Proof. $\left|f(x)-1\right|<\varepsilon$ if and only if $\left|e^x-1\right|<\varepsilon$.
$$-\varepsilon<e^x-1<\varepsilon$$
$$\ln\left(-\varepsilon+1\right)<x<\ln\left(\varepsilon+1\right)$$
$\ln\left(1-\varepsilon\right)$ is not defined for $\varepsilon\geq1$, so set $0<\varepsilon<1$. Let $\ln\left(1-\varepsilon\right)=-\delta_1$ and $\ln\left(1+\varepsilon\right)=\delta_2$. Then $\delta=\min\left(\delta_1,\delta_2\right)$. Assume $-\ln\left(1-\varepsilon\right)>\ln\left(1+\varepsilon\right)$. Then
$$0>\ln\left(1-\varepsilon\right)+\ln\left(1+\varepsilon\right)\Rightarrow0>\ln\left(\left(1-\varepsilon\right)\left(1+\varepsilon\right)\right)\Rightarrow0>\ln\left(1-\varepsilon^2\right)$$
Since $0<\varepsilon<1$, $\ln\left(1-\varepsilon^2\right)<0\Rightarrow\min\left(\delta_1,\delta_2\right)=\ln\left(1+\varepsilon\right)$. Let $\delta=\ln\left(1+\varepsilon\right)$. Then
$$-\ln\left(1+\varepsilon\right)<x<\ln\left(1+\varepsilon\right)\Rightarrow e^{-\ln\left(1+\varepsilon\right)}<e^x<e^{\ln\left(1+\varepsilon\right)}\Rightarrow\frac{1}{1+\varepsilon}<e^x<\varepsilon+1$$
$$\frac{1}{1+\varepsilon}-1<e^x-1<\varepsilon$$
Since $\frac{1}{1+\varepsilon}-1<\varepsilon$, $\left|e^x-1\right|<\varepsilon$. $\square$
 A: You appear to have all the right elements, though you got a little bit lost and confused in the middle of your proof (or at least I did). ..
You started with the right idea:  you correctly noted that if $\ln (1-\epsilon) \lt x\lt \ln (1+\epsilon )$ then we will have what we want.  
So let $\delta =\min (\ln (1+\epsilon),-\ln (1-\epsilon) )$.  This step is contained in your work, and does the trick...
A: Proof
For any $\varepsilon>0$, we take $\delta=\ln(1+\varepsilon)>0.$
Then, when $0<|x-0|<\delta$, namely, $-\ln(1+\varepsilon)<x<\ln(1+\varepsilon)$ and $x \neq 0$, we have $$1-\varepsilon<\frac{1}{1+\varepsilon}=e^{-\ln(1+\varepsilon)}<e^x<e^{\ln(1+\varepsilon)}=1+\varepsilon,$$ i.e. $$|e^x-1|<\varepsilon,$$ which is what we want to prove.
A: Without logarithms:
Assume that $x\le\dfrac1q$ where $q$ is an integer. Then using $e^{<0}<1$,
$$e^x-1=e^{x-1/q}e^{1/q}-1\le e^{1/q}-1=\frac{e-1}{e^{q-1/q}+e^{q-2/q}+\cdots1}<\frac{e-1}q\le\epsilon$$
can be achieved with $\delta:=\dfrac1{\left\lfloor\dfrac{e-1}\epsilon\right\rfloor}$.
Now assume that $-\dfrac1q\le x$ and
$$e^x-1=e^{x+1/q}e^{-1/q}-1\ge e^{-1/q}-1=\frac{e^{-1}-1}{e^{1-q/q}+e^{2-q/q}+\cdots1}>\frac{e^{-1}-1}{qe^{-1}}\ge-\epsilon$$
is also achieved with $\delta:=\dfrac1{\left\lfloor\dfrac{e-1}{\epsilon}\right\rfloor}$.
A: $$|e^x-1|<\epsilon\iff 1-\epsilon<e^x<1+\epsilon\iff \log(1-\epsilon)<x<\log(1+\epsilon)\iff\\|x|<\max(-\log(1-\epsilon),\log(1+\epsilon))=:\delta$$
as we know that these logarithms have opposite signs, and the logarithm is a monotonous function.

Notice that if $\epsilon\ge1$, the lower bound $1-\epsilon<e^x$ is always true and we simply need
$$|e^x-1|<\epsilon\iff e^x<1+\epsilon\iff x<\log(1+\epsilon)\iff\\|x|<\log(1+\epsilon)=:\delta.$$
A: Your job is easy if you know the inequality $e^x\geq 1+x$ for all $x$. Based on this inequality we have for $|x|<1$ $$x\leq e^x-1\leq \frac{x} {1-x}$$ or $$|e^x-1|\leq \max\left(|x|, \frac{|x|} {1-x}\right)$$ Now it's easy to see that $\delta =\min(1/2,\epsilon/2)$ works. 
