# How do you prove the statement lim as x approaches 1 for (5x^2)=5 using the epsilon and delta definition?

Oct 25, 2015

See the explanation section below.

#### Explanation:

Preliminary Analysis

We need to make $\left\mid 5 {x}^{2} - 5 \right\mid < \epsilon$ by making $\left\mid x - 1 \right\mid < \delta$. (We choose the $\delta$.)

So we begin by examining $\left\mid 5 {x}^{2} - 5 \right\mid$

$\left\mid 5 {x}^{2} - 5 \right\mid = \left\mid 5 \left({x}^{2} - 1\right) \right\mid = \left\mid 5 \right\mid \left\mid {x}^{2} - 1 \right\mid$

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

We control, through our choice of $\delta$, the maximum for $\left\mid x - 1 \right\mid$.

If we make sure that $\delta$ is less than a number we choose now, then we can also control the size of $\left\mid x + 1 \right\mid$.
(If we put a bound on the distance between $x$ and $1$, the we also get a bound on the distance between $x$ and $- 1$)

Any number will do, but is might be easiest for purpose of illustration to pick a number we haven't use yet so we can keep track of it.

Let's make sure that $\delta \le 2$.

If $\left\mid x - 1 \right\mid < 2$, then $- 2 < x - 1 < 2$.

So, adding 2 to each part, we get, $0 < x + 1 < 4$ and $\left\mid x + 1 \right\mid < 4$.

We want $5 \left\mid x + 1 \right\mid \left\mid x - 1 \right\mid < \epsilon$ and we plan to make sure that $\left\mid x + 1 \right\mid < 4$, so we now need

$5 \left\mid x + 1 \right\mid \left\mid x - 1 \right\mid$

$\text{which is} < 5 \left(4\right) \left\mid x - 1 \right\mid$
$\text{which is} = 20 \left\mid x - 1 \right\mid$
$\text{we want this} < \epsilon$

So let's make sure that, in addition to $\delta \le 2$,

we also want $\delta \le \frac{\epsilon}{20}$

Now we are ready to write the proof.

Proof
Claim: ${\lim}_{x \rightarrow 1} 5 {x}^{2} = 5$

Given $\epsilon > 0$, let $\delta = \min \left\{2 , \frac{\epsilon}{20}\right\}$ (Note that $\delta > 0$ as required.)

If $x$ is chosen so that $0 < \left\mid x - 1 \right\mid < \delta$ then

note first that $\left\mid x + 1 \right\mid < 4$
( $\left\mid x - 1 \right\mid < \delta \le 2 \Rightarrow - 2 < x - 1 < 2 \Rightarrow 0 < x + 1 < 4$).

Furthermore, for such $x$, we have

$\left\mid 5 {x}^{2} - 5 \right\mid = 5 \left\mid {x}^{2} - 1 \right\mid = 5 \left\mid x + 1 \right\mid \left\mid x - 1 \right\mid$

$< 5 \left(4\right) \left\mid x - 1 \right\mid = 20 \left\mid x - 1 \right\mid$

$< 20 \left(\frac{\epsilon}{20}\right) = \epsilon$.

That is: for 0 < abs(x-1< delta, we have #abs(5x^2-5) < epsilon.

Therefore, by the definition on limit,

${\lim}_{x \rightarrow 1} 5 {x}^{2} = 5$