Spivak tough limit proof-verification Suppose there is a $\delta > 0$ such that $f(x) = g(x)$ when $0 < |x - a| < \delta$. Prove that $\displaystyle \lim_{x\to a} f(x) = \lim_{x \to a} g(x)$.
$|f(x) - L| < \epsilon$ for $|x - a| < \delta_1$ 
$|g(x) - M| < \epsilon$ for $|x - a| <\delta_2$ 
Let $\delta' = \min(\delta_1, \delta_2)$ 
Let $\delta' < \delta$    
Therefore, $f(x) = g(x)$ for this interval $\delta'$ 
This gives us: 
$|f(x) - L| < \epsilon$ for $|x - a| < \delta_1$ 
$|f(x) - M| < \epsilon$ for $|x - a| <\delta_2$ 
Thus, what is required is to prove $M = L$ 
Assume $M \ne L$ 
Since $\epsilon \in (0, \infty)$, this means $\epsilon = |L - M|/2$ is a possibility. 
$|f(x) - L| < |L - M|/2$ for $|x - a| < \delta_1$ 
$|f(x) - M| < |L - M|/2$ for $|x - a| <\delta_2$ 
$|L - M| = |-(f(x) - L) + (f(x) - M)|$
The triangle inequality states,
$|f(x) - L| + |f(x) - M| \ge |(f(x) - M) - (f(x) - L)|$
Then, 
$ |L - M|/2 + |L - M|/2 > |f(x) - L| + |f(x) - M| \ge |(f(x) - M) - (f(x) - L)|$
Which is a contradiction, therefore $L \ne M$ is false, and $L = M$ is the true statement. 
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 A: Let $L = \lim_{x\to a} f(x)$ and $M = \lim_{x\to a} g(x)$. Let $\epsilon > 0$. Choose positive numbers $\delta_1$ and $\delta_2$ such that $|f(x) - L| < \epsilon/2$ whenever $0 < |x - a| < \delta_1$, and $|g(x) - M| < \epsilon/2$ whenever $0 < |x - a| < \delta_2$. Set $\delta_3 = \min\{\delta_1, \delta_2\}$. For all $x$, $0 < |x - a| < \delta_3$ implies 
\begin{equation}
|L - M| \le |L - f(x)| + |f(x) - g(x)| + |g(x) - M| = |L - f(x)| + |g(x) - M| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
\end{equation}
Since $\epsilon$ was arbitrary, $L = M$.
A: I think it is much better to appeal to the definition of limits. A simple answer for the simple question is as follows:
Limit of a function $f$ at point $a$ is defined in terms of values of $f$ in a deleted neighborhood of $a$. Hence if there is another function $g$ whose values are same as that of $f$ in a deleted neighborhood of $a$ then the behavior of both $f$ and $g$ is same as far as limit at point $a$ is concerned.
It does not make sense to invoke $\epsilon$'s for every trivial argument/inference.
A: The proof looks okay!
I think, however, that you would need to assume that the limits themselves exist first, right? Otherwise you could get something like $\lim_{x\rightarrow 0 } \frac{1}{x} = \lim_{x\rightarrow 0 } \frac{1}{x} $, which is not true. It's just a formalism, though.
By the way, if you consider the function $(f-g)(x)$, it is easy to verify that $\lim_{x\rightarrow a } (f-g)(x) = 0$. If $\lim_{x\rightarrow a } f(x)$ exists, then $\lim_{x\rightarrow a } g(x)=\lim_{x\rightarrow a } f(x)+\lim_{x\rightarrow a } (f-g)(x)$. It's way faster! Your solution looks fine though.
A: One can also do it easily with sequences. Let $x_n$ be a sequence such that $x_n \neq a$ and $x_n\to a$. Per definition there is some $N$ such that for all $n>N$ we have $0<|x_n-a|<\delta$. So for all $n>N$ we have $f(x_n)=g(x_n) $. Hence the limit if $n\to\infty$ is the same if it exists (this assumption is important!) . Since $x_n$ was arbitrary the claim follows. 
