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:


 * $\ds \lim_{x \mathop \to c} \map f x$

exist.

Then the limit:


 * $\ds \lim_{x \mathop \to c} \map g x$

also exists, and:


 * $\ds \lim_{x \mathop \to c} \map f x = \lim_{x \mathop \to c} \map g x$

Proof
By hypothesis:


 * $\map f x = \map g x$

for all $x$ such that:


 * $x \in \R: 0 < \size {\alpha - x} < \epsilon$

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


 * $x \in \R: 0 < \size {x - c} < \delta$

By definition, that the limit of $\map f x$ exists is to say:


 * $\exists L: \forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$

For all $x$ relevant, we can substitute $\map f x$ with $\map g x$ and get:


 * $\exists L: \forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map g x - L} < \epsilon$

The result follows by the definition of limit.