# How do you prove the statement lim as x approaches 2 for  (x^2 - 3x) = -2 using the epsilon and delta definition?

Oct 19, 2015

See the explanation, below.

#### Explanation:

Preliminary Analysis

We want to make $\left\mid \left({x}^{2} - 3 x\right) - \left(- 2\right) \right\mid < \epsilon$

By choosing $\delta$, we control the maximum size of $\left\mid x - 2 \right\mid$

We now look at $\left\mid \left({x}^{2} - 3 x\right) - \left(- 2\right) \right\mid$ keeping in mind what we control.

$\left\mid \left({x}^{2} - 3 x\right) - \left(- 2\right) \right\mid = \left\mid {x}^{2} - 3 x + 2 \right\mid$

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

$= \left\mid x - 1 \right\mid \left\mid x - 2 \right\mid$

If we knew the size of $\left\mid x - 1 \right\mid$, we could choose $\left\mid x - 2 \right\mid$ to make sure that the product is $< \epsilon$

Let's start by making sure that $x$ is a little close to $2$, say $\left\mid x - 2 \right\mid < 1$ ($\delta$ will need to be at most $1$.)

If $\left\mid x - 2 \right\mid < 1$, then $- 1 < x - 2 < 1$, that is: $1 < x < 3$.

If follows that $0 < x - 1 < 2$, so we have $\left\mid x - 1 \right\mid < 2$

Now if we ALSO make sure that $\left\mid x - 2 \right\mid < \frac{\epsilon}{2}$,

then we will have $\left\mid x - 1 \right\mid \left\mid x - 2 \right\mid < 2 \left\mid x - 2 \right\mid < 2 \left(\frac{\epsilon}{2}\right) = \epsilon$

Now we are ready to write the proof:

Proof

Given $\epsilon > 0$, let $\delta = \min \left\{1 , \frac{\epsilon}{2}\right\}$.
(Note that #delta is positive.)

For every $x$ that satisfies $0 < \left\mid x - 2 \right\mid < \delta$
we have

($\left\mid x - 1 \right\mid < 2$ and also)

$\left\mid \left({x}^{2} - 3 x\right) - \left(- 2\right) \right\mid = \left\mid {x}^{2} - 3 x + 2 \right\mid$

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

$= \left\mid x - 1 \right\mid \left\mid x - 2 \right\mid$

$< \left(2\right) \left(\frac{\epsilon}{2}\right) = \epsilon$

We have shown that for any $\epsilon > 0$ there is a $\delta > 0$ such that

if $0 < \left\mid x - 2 \right\mid < \delta$, then $\left\mid \left({x}^{2} - 3 x\right) - \left(- 2\right) \right\mid < \epsilon$.

By the definition of limit, ${\lim}_{x \rightarrow 2} \left({x}^{2} - 3 x\right) = - 2$.