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 $\frac{1}{1+\varepsilon}-1<\varepsilon$, $\left|e^x-1\right|<\varepsilon$. $\square$
Best Answer
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.