Limit of Functions that Agree

Theorem
Let $f$ and $g$ be real functions.

Let $f$ and $g$ agree for all $x$ in a deleted neighborhood of $c$.

Let the limit:


 * $\displaystyle \lim_{x \to c} \ f \left({x}\right)$

exist.

Then the limit:


 * $\displaystyle \lim_{x \to c} \ g \left({x}\right)$

also exists, and:


 * $\displaystyle \lim_{x \to c} \ f \left({x}\right) = \lim_{x \to c} \ g \left({x}\right)$

Proof
By hypothesis:


 * $f \left({x}\right) = g \left({x}\right)$

for all $x$ such that:


 * $x \in \R: 0 < \left \vert{\alpha - x}\right \vert < \epsilon$

Putting $c$ for $\alpha$ and $\delta$ for $\epsilon$, this is equivalent to:


 * $x \in \R: 0 < \left \vert{x - c}\right \vert < \delta$

By definition, that the limit of $f \left({x}\right)$ exists is to say:


 * $\exists L: \forall \varepsilon > 0: \exists \delta > 0: 0 < \left \vert{x - c}\right \vert < \delta \implies \left \vert{f \left({x}\right) - L}\right \vert < \varepsilon$

For all $x$ relevant, we can substitute $f \left({x}\right)$ with $g \left({x}\right)$ and get:


 * $\exists L: \forall \varepsilon > 0: \exists \delta > 0: 0 < \left \vert{x - c}\right \vert < \delta \implies \left \vert{g \left({x}\right) - L}\right \vert < \varepsilon$

The result follows by the definition of limit.