# How do you prove the statement lim as x approaches 6 for ((x/4)+3) = 9/2 using the epsilon and delta definition?

Apr 8, 2016

Please see the explanation section below.

#### Explanation:

Preliminary analysis

${\lim}_{x \rightarrow \textcolor{g r e e n}{a}} \textcolor{red}{f \left(x\right)} = \textcolor{b l u e}{L}$ if and only if

for every $\epsilon > 0$, there is a $\delta > 0$ such that:
for all $x$, $\text{ }$ if $0 < \left\mid x - \textcolor{g r e e n}{a} \right\mid < \delta$, then $\left\mid \textcolor{red}{f \left(x\right)} - \textcolor{b l u e}{L} \right\mid < \epsilon$.

So we want to make $\left\mid {\underbrace{\textcolor{red}{\left(\frac{x}{4} + 3\right)}}}_{\textcolor{red}{f \left(x\right)}} - {\underbrace{\textcolor{b l u e}{\frac{9}{2}}}}_{\textcolor{b l u e}{L}} \right\mid$ less than some given $\epsilon$ and we control (through our control of $\delta$) the size of $\left\mid x - {\underbrace{\textcolor{g r e e n}{6}}}_{\textcolor{g r e e n}{a}} \right\mid$

Look at the thing we want to make small:

$\left\mid \left(\frac{x}{4} + 3\right) - \frac{9}{2} \right\mid = \left\mid \frac{x}{4} - \frac{3}{2} \right\mid = \left\mid \frac{x - 6}{4} \right\mid = \frac{\left\mid x - 6 \right\mid}{\left\mid 4 \right\mid} = \frac{\left\mid x - 6 \right\mid}{4}$

And there's the thing we control, in the numerator!

We can make $\frac{\left\mid x - 6 \right\mid}{4} < \epsilon$ by making $\left\mid x - 6 \right\mid < 4 \epsilon$.

So we will choose $\delta = 4 \epsilon$. (Any lesser $\delta$ would also work.)

(Detail: if $\left\mid x - 6 \right\mid < 4 \epsilon$, then we can multiply on both sides by the positive number $\frac{1}{4}$ to get $\frac{\left\mid x - 6 \right\mid}{4} < \epsilon$.)

Now we need to actually write up the proof:

Proof

Given $\epsilon > 0$, choose $\delta = 4 \epsilon$. $\text{ }$ (note that $\delta$ is also positive).

Now for every $x$ with $0 < \left\mid x - 6 \right\mid < \delta$, we have

$\left\mid f \left(x\right) - \frac{9}{2} \right\mid = \left\mid \left(\frac{x}{4} + 3\right) - \frac{9}{2} \right\mid = \left\mid \frac{x - 6}{4} \right\mid = \frac{\left\mid x - 6 \right\mid}{4} < \frac{\delta}{4}$

[Detail if $\left\mid x - 6 \right\mid < \delta$, we can conclude that $\frac{\left\mid x - 6 \right\mid}{4} < \frac{\delta}{4}$. -- we usually do not mention this, but leave it to the reader. See below.]

And $\frac{\delta}{4} = \frac{4 \epsilon}{4} = \epsilon$

Therefore, with this choice of delta, whenever $0 < \left\mid x - 6 \right\mid < \delta$, we have $\left\mid f \left(x\right) - \frac{9}{2} \right\mid < \epsilon$

So, by the definition of limit, ${\lim}_{x \rightarrow 6} \left(\frac{x}{4} + 3\right) = \frac{9}{2}$.

We can condense a bit

for every $x$ with $0 < \left\mid x - 6 \right\mid < \delta$, we have

$\left\mid f \left(x\right) - \frac{9}{2} \right\mid = \left\mid \left(\frac{x}{4} + 3\right) - \frac{9}{2} \right\mid$

$= \left\mid \frac{x - 6}{4} \right\mid$

$= \frac{\left\mid x - 6 \right\mid}{4}$

$< \frac{\delta}{4} = \frac{4 \epsilon}{4} = \epsilon$.

So, $\left\mid f \left(x\right) - \frac{9}{2} \right\mid < \epsilon$.