# How do you prove the statement lim as x approaches 3 for (x/5) = 3/5 using the epsilon and delta definition?

Oct 10, 2015

Preliminary work
Recall the definition:
${\lim}_{\textcolor{red}{x \rightarrow c}} \textcolor{b l u e}{f} \left(x\right) = \textcolor{b l u e}{L}$

If and only if

for every positive number $\textcolor{b l u e}{\epsilon}$ , there is a positive number $\textcolor{red}{\delta}$ for which the following is true:
If $0 < \left\mid \textcolor{red}{x - c} \right\mid < \textcolor{red}{\delta}$, then $\left\mid \textcolor{b l u e}{f \left(x\right) - L} \right\mid < \textcolor{b l u e}{\epsilon}$,

To show ${\lim}_{\textcolor{red}{x \rightarrow 3}} \textcolor{b l u e}{\frac{x}{5}} = \textcolor{b l u e}{\frac{3}{5}}$ we need to show our reader that

for every positive number $\textcolor{b l u e}{\epsilon}$ , there is a positive number $\textcolor{red}{\delta}$ for which the following is true:
If $0 < \left\mid \textcolor{red}{x - 3} \right\mid < \textcolor{red}{\delta}$, then $\left\mid \textcolor{b l u e}{\frac{x}{5} - \frac{3}{5}} \right\mid < \textcolor{b l u e}{\epsilon}$,

We want to make $\left\mid \textcolor{b l u e}{\frac{x}{5} - \frac{3}{5}} \right\mid$ smaller than $\textcolor{b l u e}{\epsilon}$, and we control (through $\textcolor{red}{\delta}$) the size of $\left\mid \textcolor{red}{x - 3} \right\mid$

Notice that
$\left\mid \textcolor{b l u e}{\frac{x}{5} - \frac{3}{5}} \right\mid = \left\mid \textcolor{b l u e}{\frac{1}{5} \left(x - 3\right)} \right\mid$

$= \textcolor{b l u e}{\frac{1}{5}} \left\mid \textcolor{b l u e}{\left(x - 3\right)} \right\mid$

$= \frac{1}{5} \left\mid \textcolor{red}{\left(x - 3\right)} \right\mid$

So, if we make $\left\mid \textcolor{red}{\left(x - 3\right)} \right\mid < 5 \textcolor{b l u e}{\epsilon}$, then we will have the desired result, because

$\left\mid \textcolor{red}{\left(x - 3\right)} \right\mid < 5 \textcolor{b l u e}{\epsilon}$ implies that $\frac{1}{5} \left\mid \textcolor{red}{\left(x - 3\right)} \right\mid < \frac{1}{5} \left(5 \textcolor{b l u e}{\epsilon}\right)$ which is equal to color(blue)(epsilon)).

Now, we are finished with our preliminary considerations and are ready to present our proof to the world.

(I'll keep the colors here to help us understand.)
Note also that saying a number is positive is the same as saying that it is $> 0$

Proof that ${\lim}_{\textcolor{red}{x \rightarrow 3}} \textcolor{b l u e}{\frac{x}{5}} = \textcolor{b l u e}{\frac{3}{5}}$

Given $\textcolor{b l u e}{\epsilon} > 0$ , choose $\textcolor{red}{\delta} = 5 \textcolor{b l u e}{\epsilon}$

Now if $0 < \left\mid \textcolor{red}{x - 3} \right\mid < \textcolor{red}{\delta}$, then we have:

$\left\mid \textcolor{b l u e}{\frac{x}{5} - \frac{3}{5}} \right\mid = \left\mid \textcolor{b l u e}{\frac{1}{5} \left(x - 3\right)} \right\mid$

$= \textcolor{b l u e}{\frac{1}{5}} \left\mid \textcolor{b l u e}{\left(x - 3\right)} \right\mid$

$= \frac{1}{5} \left\mid \textcolor{red}{\left(x - 3\right)} \right\mid$

$> \frac{1}{5} \left(5 \textcolor{b l u e}{\epsilon}\right)$

$= \textcolor{b l u e}{\epsilon}$

That is:
If $0 < \left\mid \textcolor{red}{x - 3} \right\mid < \textcolor{red}{\delta}$, then $\left\mid \textcolor{b l u e}{\frac{x}{5} - \frac{3}{5}} \right\mid < \textcolor{b l u e}{\epsilon}$.

So, by the definition of limit, we conclude that

${\lim}_{x \rightarrow 3} \left(\frac{x}{5}\right) = \frac{3}{5}$