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 [1].

*Essential Background Information:*

Definition of a Limit:

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:*

[1]

https://tutorial.math.lamar.edu/classes/calci/limitproofs.aspx
[2] David V. Widder. Advanced Calculus. Dover 1989.

*Related*