Proving that limits are a "local property" From M. Spivak's Calculus, let me reproduce Problem 10 in the chapter Limits.

Suppose there is a $\delta>0$ such that $f(x) = g(x)$ when $0<|x-a|<\delta$. Prove that $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$. In other words, $\lim_{x\to a}f(x)$ depends only on the values of $f(x)$ near $a$ — this fact is often expressed by saying that limits are a "local property".

I would like help identifying whether my proof is correct.
Claim: Given that there exists $\delta'$ such that when $0<|x-a|<\delta'$, $f(x) = g(x)$, show that $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$.
Proof: Since by assumption $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist, we may restate the problem as needing to show that $\lim_{x\to a}\big(f(x)-g(x)\big)=0$. 
Hence, given $\epsilon>0$, take $\delta:=\delta'$ so that if $0<|x-a|<\delta$, $|f(x)-g(x)-0| = |f(x)-g(x)|=0$. 
This implies by definition that $|f(x)-g(x)-0|<\epsilon$.
$$\tag*{$\blacksquare$}$$
 A: This is very easy using for example, the sequential criteria: 

$\lim_{x\to a}f(x)=L$ if and only if $\forall (x_n)$, so that $(x_n)\neq a,\forall n\in\mathbb{N}$ and $(x_n)\rightarrow a$ then it follows that $f(x_n)\rightarrow L$.

With this criteria, given $(x_n)\neq a,\forall n\in\mathbb{N}$ so that $(x_n)\rightarrow a$ and suppose $\lim_{x\to a}f(x)=L$, then given $\delta>0$ as in your question, for the definition of sequence limit, $\exists N\in\mathbb{N}$ so that $|x_n-a|<\delta,\forall n>N$
Then by hypothesis, $f(x_{n+N})=g(x_{n+N})$ and so $L=\lim f(x_n)=\lim f(x_{n+N})=\lim g(x_{n+N})=\lim g(x_n)$
A: What you have shown is the following:

If $\lim \limits_{x \to a}f(x)$ and $\lim \limits_{x \to a}g(x)$ exist, and there is a $\delta'>0$ such that $f(x) = g(x)$ whenever $0<|x-a|<\delta'$, then $\lim \limits_{x \to a}f(x) = \lim \limits_{x \to a}g(x)$.

Your proof of this assertion seems fine.
As suggested in the comments, you can also prove the following statement (or a similar statement with the roles of $f$ and $g$ reversed): 

Suppose $\lim \limits_{x \to a}f(x)$ exists, and there is a $\delta' > 0$ such that $f(x) = g(x)$ whenever $0 < |x-a| < \delta'$. Then, $\lim \limits_{x \to a}g(x)$ also exists, and $\lim \limits_{x \to a}f(x) = \lim \limits_{x \to a}g(x)$.

The proof of this is very similar to the one you provided. Denote $l:= \lim \limits_{x \to a}f(x)$. Now, let $\varepsilon> 0$ be given. Then, we know there exists a $\delta_1 > 0$ such that if $0 < |x-a| < \delta_1$ then $|f(x) -l| < \varepsilon$. Now, define $\delta := \min(\delta', \delta_1)$. Now, if $0< |x-a|< \delta$ then we have that
\begin{align}
|g(x) - l| &= |f(x) - l| \\
&< \varepsilon.
\end{align}
Since $\varepsilon>0$ is arbitrary, this shows $\lim \limits_{x \to a}g(x)$ exists and equals $l$ as well.
