I'm trying to prove the following result.
Let $f$ be a function. Then:
$$\lim_{x \to x_0} f(x) = c \iff f(x) = c + \alpha(x) \land \lim_{x \to x_0} \alpha(x) = 0$$
where $c \in \mathbb{R}$.
Proof Attempt:
We prove the forward conditional first. Let $\lim_{x \to x_0} f(x) = c$. Then, for any $\epsilon > 0$:
$$\exists \delta > 0: 0 < |x-x_0| < \delta \implies |f(x) - L| < \epsilon$$
Defined $\alpha(x) = f(x) - c$. So, we have $f(x) = c + \alpha(x)$. Now, we have to prove that $\lim_{x \to x_0} \alpha(x) = 0$.
Let $\epsilon > 0$. Then, we have to show that there exists a $\delta > 0$ such that:
$$0 < |x-x_0| < \delta \implies |\alpha(x) - 0| = |\alpha(x)| < \epsilon$$
However, we note that:
$$|\alpha(x)| = |f(x) - c| < \epsilon$$
And we know that such a deleted neighbourhood of $x_0$ exists so that the inequality above holds. So, we have proven that $\lim_{x \to x_0} \alpha(x) = 0$.
Now, we prove the backwards conditional. Let $f(x) = c + \alpha(x)$, where $\lim_{x \to x_0} \alpha(x) = 0$. For any $\epsilon > 0$, we have to show that there exists a $\delta > 0$ such that:
$$0 < |x-x_0| < \delta \implies |f(x) - c| < \epsilon$$
However, we note that:
$$|f(x)-c| = |[\alpha(x)+c]-c| = |\alpha(x)| < \epsilon$$
And, by our hypothesis, there does exist a $\delta$ such that the inequality above holds. Hence, we have shown that $\lim_{x \to x_0} f(x) = c$.
This proves the desired assertion.
Is my proof above correct? This seemed easy, so I'm wondering if I'm being an idiot and being deceived by the apparent simplicity of the problem. If it isn't correct, how can I fix the proof?
Suppose $\lim_{x\rightarrow x_0 } f(x) = c$ then $\lim_{x\rightarrow x_0 } f(x) -c = 0$. now define $\alpha(x)=f(x)-c$. So that gives you $\lim_{x\rightarrow x_0 } \alpha(x) = 0$
On the other direction, you follow the proof from end to stard.
This statement sometimes used as definition and it is a very basic one, so it makes sense that the proof is easy.