# How do you prove that the limit of 5 - 2x= 1 as x approaches 2 using the epsilon delta proof?

Apr 25, 2016

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 $\left\mid {\underbrace{\textcolor{red}{\left(5 - 2 x\right)}}}_{\textcolor{red}{f \left(x\right)}} - {\underbrace{\textcolor{b l u e}{1}}}_{\textcolor{b l u e}{L}} \right\mid$ less than some given $\epsilon$ and we control (through our control of $\delta$) the size of $\left\mid x - {\underbrace{\textcolor{g r e e n}{2}}}_{\textcolor{g r e e n}{a}} \right\mid$

Look at the thing we want to make small. We want to see the thing we control.

$\left\mid \left(5 - 2 x\right) - 1 \right\mid = \left\mid - 2 x + 4 \right\mid = \left\mid - 2 \left(x - 2\right) \right\mid = \left\mid - 2 \right\mid \left\mid x - 2 \right\mid = 2 \left\mid x - 2 \right\mid$

And there's $\left\mid x - 2 \right\mid$, the thing we control

We can make $2 \left\mid x - 2 \right\mid < \epsilon$ by making $\left\mid x - 2 \right\mid < \frac{\epsilon}{2}$.

So we will choose $\delta = \frac{\epsilon}{2}$. (Any lesser $\delta$ would also work.)

(Detail: if $\left\mid x - 2 \right\mid < \frac{\epsilon}{2}$, then we can multiply on both sides by the positive number $2$ to get $2 \left\mid x - 2 \right\mid < \epsilon$.)

Now we need to actually write up the proof:

Proof

Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{2}$. $\text{ }$ (note that $\delta$ is also positive).

Now for every $x$ with $0 < \left\mid x - 2 \right\mid < \delta$, we have

abs(f(x)-1) = abs((5-2x) - 1) = abs(-2x+4)) = abs(-2)abs(x-2) = 2abs(x-2) < 2delta

[Detail if $\left\mid x - 2 \right\mid < \delta$, we can conclude that $2 \left\mid x - 2 \right\mid < 2 \delta$. $\text{ }$ We usually do not mention this, but leave it to the reader. See below.]

And $2 \delta = 2 \frac{\epsilon}{2} = \epsilon$

Therefore, with this choice of delta, whenever $0 < \left\mid x - 2 \right\mid < \delta$, we have $\left\mid f \left(x\right) - 1 \right\mid < \epsilon$

So, by the definition of limit, ${\lim}_{x \rightarrow 2} \left(5 - 2 x\right) = 1$.

We can condense a bit

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

$\left\mid f \left(x\right) - 1 \right\mid = \left\mid \left(5 - 2 x\right) - 1 \right\mid$

$= \left\mid - 2 \left(x - 2\right) \right\mid$

$= 2 \left\mid x - 2 \right\mid$

$< 2 \delta = 2 \frac{\epsilon}{2} = \epsilon$.

So, $\left\mid f \left(x\right) - 1 \right\mid < \epsilon$.