Proving $\lim_{x\to 0} \frac{e^x -1}{x} =1$ using epsilon delta and the continuity of $e^x$ at $0$.
Knowing that $e^x$ is continous at 1, I want to proof the limit $\lim_{x\to 0}\frac{e^x -1}{x} =1$ using an epsilon-delta proof. I reasoned as follows:
Let $\varepsilon$ be greater than 0 an suppose that $$\left|\frac{e^x-1}{x} -1\right|<\varepsilon$$
This means that $$-\varepsilon -1 < -\varepsilon +1 < \frac{e^x -1}{x} < \varepsilon +1$$ Then,
$$\frac{|e^x-1|}{|x|}=\left|\frac{e^x-1}{x}\right|<\varepsilon+1$$. So, if I multiply both sides by |x| we have that $$|e^x -1|<|x|(\varepsilon+1)$$
Let $\varepsilon_2 = |x|(\varepsilon +1)$. Since we know that $e^x$ is continous at x=1 we then have that there exists $\delta$ such that $$0<|x|<\delta \Rightarrow |e^x -1|<\varepsilon_2$$ which then, by replacing the definition of $\varepsilon_2$ implies $$0<|x|<\delta \Rightarrow\left|\frac{e^x-1}{x} -1\right|<\varepsilon$$
Although it might be convincing, I'm having a hard time to check whether the step of naming $\varepsilon_2=|x|(\varepsilon+1)$ is indeed valid for the proof.
First of all by Archimedian Property we know that for some $\epsilon_2$=$|x|$ $\exists N\in\mathbb{N}$ such that $\frac{1}{N}<\epsilon_2 \Rightarrow |x|>\frac{1}{N_1}$ after that we need to get the inequality. Pick $\delta :=min(1,N_1)$ and $\epsilon_1 = eN_1+N_1+1$ after that we have two inequalities $-1<x<1 \Rightarrow e>e^x>0$ and which comes from Archimeadian property. Then obtain; $|\frac{e^x-1}{x}-1|\leq \frac{|e^x-1|}{|x|}+1\leq (e^x+1).N_1+1\leq\ eN_1+N_1+1\leq \epsilon_1 $ then we are done.