# How do you prove the statement lim as x approaches 2 for (x^2 - 4x + 5) = 1 using the epsilon and delta definition?

Mar 28, 2017

#### Explanation:

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.

Finding the proof

By definition,

${\lim}_{x \rightarrow \textcolor{g r e e n}{a}} \textcolor{red}{f \left(x\right)} = \textcolor{b l u e}{L}$ 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 $\left\mid \textcolor{red}{f \left(x\right)} - \textcolor{b l u e}{L} \right\mid < \epsilon$.

We have been asked to show that

${\lim}_{x \rightarrow \textcolor{g r e e n}{2}} \textcolor{red}{{x}^{2} - 4 x + 5} = \textcolor{b l u e}{1}$

So we want to make $\left\mid {\underbrace{\textcolor{red}{{x}^{2} - 4 x + 5}}}_{\textcolor{red}{f \left(x\right)}} - {\underbrace{\textcolor{b l u e}{1}}}_{\textcolor{b l u e}{L}} \right\mid$ less than some given $\epsilon$ and we control (through our control of $\delta$) the size of $\left\mid x - {\underbrace{\textcolor{g r e e n}{\left(2\right)}}}_{\textcolor{g r e e n}{a}} \right\mid$

We want: $\left\mid \left({x}^{2} - 4 x + 5\right) - 1 \right\mid < \epsilon$

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

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

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

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

In order to make this less than $\epsilon$, it suffices to make $\left\mid x - 2 \right\mid$ less than $\sqrt{e} \psi$

Proving our L is correct -- Writing the proof

Claim: ${\lim}_{x \rightarrow 2} \left({x}^{2} - 4 x + 5\right) = 1$

Proof:

Given $\epsilon > 0$, choose $\delta = \sqrt{\epsilon}$. (Note that $\delta$ is positive.)

Now if $0 < \left\mid x - 2 \right\mid < \delta$ then

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

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

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

$< {\delta}^{2}$ $\text{ }$ (See Note below)

$= {\left(\sqrt{\epsilon}\right)}^{2}$

$= \epsilon$

We have shown that for any positive $\epsilon$, there is a positive $\delta$ such that for all $x$, if $0 < \left\mid x - 2 \right\mid < \delta$, then $\left\mid \left({x}^{2} - 4 x + 5\right) - 1 \right\mid < \epsilon$.

So, by the definition of limit, we have ${\lim}_{x \rightarrow 2} \left({x}^{2} - 4 x + 5\right) = 1$.

Note

Since the squaring function is increasing on positive values,

$\left\mid x - 2 \right\mid < \sqrt{\delta}$ implies that ${\left\mid x - 2 \right\mid}^{2} = {\left(x - 2\right)}^{2} < {\delta}^{2}$