# How do you prove the statement lim as x approaches 2 for ((x^2+x-6)/(x-2))=5 using the epsilon and delta definition?

Oct 16, 2015

See the explanation below.

#### Explanation:

Preliminary Analysis

First look at $\left\mid f \left(x\right) - L \right\mid$. I want to make that less than $\epsilon$.

I am looking for $\left\mid x - 2 \right\mid$ because, through my control of $\delta$, I control $\left\mid x - 2 \right\mid$.

$\left\mid f \left(x\right) - L \right\mid = \left\mid \frac{{x}^{2} + x - 6}{x - 2} - 5 \right\mid$

$= \left\mid \frac{\left(x + 3\right) \left(x - 2\right)}{x - 2} - 5 \right\mid$

$= \left\mid \left(x + 3\right) - 5 \right\mid$ for all $x \ne 2$

$= \left\mid x - 2 \right\mid$ for all $x \ne 2$

We want to make $\left\mid f \left(x\right) - L \right\mid$ less than a given $\epsilon$.

If we make $\left\mid x - 2 \right\mid < \epsilon$ BUT $x \ne 2$, then we will have what we want. (We cannot allow $x = 2$ because $2$ is not in the domain of $f$.

So now, we are ready to write the proof.

Proof

Given $\epsilon > 0$ choose $\delta = \epsilon$. (Clearly, then, $\delta > 0$ as required.

Now if $x$ is chosen so that

$0 < \left\mid x - 2 \right\mid < \delta$, then we will have

$\left\mid \frac{{x}^{2} + x - 6}{x - 2} - 5 \right\mid = \left\mid \frac{\left(x + 3\right) \left(x - 2\right)}{x - 2} - 5 \right\mid$

$= \left\mid \left(x + 3\right) - 5 \right\mid = \left\mid x - 2 \right\mid < \delta = \epsilon$.

That is:
If $0 < \left\mid x - 2 \right\mid < \delta$, then $\left\mid \frac{{x}^{2} + x - 6}{x - 2} - 5 \right\mid < \epsilon$.

By the definition of limit,

${\lim}_{x \rightarrow 2} \frac{{x}^{2} + x - 6}{x - 2} = 5$