In this post, I prove that

$\displaystyle \lim_{x \rightarrow a} c = c$

if $c$ is a variable for any real number. To do this, I approximately follow the outline from reference .

Essential Background Information:

A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and only if, for each positive number $\epsilon$ there is another, $\delta$, such that whenever $0 < |x-a| < \delta$ we have $|f(x) – A| < \epsilon$. That is, when $x$ is near $a$ (within a distance $\delta$ from it), $f(x)$ is near $A$ (within a distance $\epsilon$ from it). In symbols we write $\lim_{x \rightarrow a} f(x) = A$.

Objective:

The objective is to show that

$|f(x) – A| < \epsilon$ whenever $0 < |x-a| < \delta$,

which is a slight change of wording of the limit definition.

Proof:

If $\displaystyle \lim_{x \rightarrow a} c = c$ is correct, the variable $c$ takes the role of $A$ in the limit definition. So the objective is actually to show that $|f(x) – c| < \epsilon$ whenever $0 < |x-a| < \delta$. For the case of $\displaystyle \lim_{x \rightarrow a} c$, the variable $c$ also takes the role of $f(x)$ in the limit definition. Hence $|f(x) – A|$ simplifies to $|f(x) – c|$ which reduces to $|c – c|$ which simplifies to $0$. In short, $0 = |f(x) – A|$. The limit definition requires $0 < \epsilon$, so by substituting $0 = |f(x) – A|$ it is clear that $|f(x) – A| < \epsilon$ is true for any allowed value of $\delta$ as there is no functional relation between $\epsilon$ and $\delta$. Thus, whenever $0 < |x-a| < \delta$, $|f(x) – A| < \epsilon$. $\quad \square$

References:

 https://tutorial.math.lamar.edu/classes/calci/limitproofs.aspx

 David V. Widder. Advanced Calculus. Dover 1989.