# How do you use the epsilon delta definition to find the limit of ((x^2+x-6)/(x-2)) as x approaches 2?

Oct 30, 2016

The epsilon-delta definition is not useful for finding limits. It is useful for proving that the limit is what I claim it is.

#### Explanation:

To find ${\lim}_{x \rightarrow 2} \frac{{x}^{2} + x - 6}{x - 2}$, we can use algebra and the following theorem:

If $f \left(x\right) = g \left(x\right)$ for all $x$ in an open interval containing $a$, except possibly at $a$, and if ${\lim}_{x \rightarrow a} g \left(x\right) = L$,

then ${\lim}_{x \rightarrow a} f \left(x\right) = L$.

(The theorem is proved using the epsilon-delta definition of limit.)

Note that $\frac{{x}^{2} + x - 6}{x - 2} = \frac{\left(x + 3\right) \left(x - 2\right)}{x - 2} = x + 3$ for all $x$ except $2$.

Furthermore, by properties of limits (which are proved using the $\epsilon - \delta$ definition) , ${\lim}_{x \rightarrow 2} \left(x + 3\right) = 5$.

Therefore, by the theorem previously cited, ${\lim}_{x \rightarrow 2} \frac{{x}^{2} + x - 6}{x - 2} = 5$.

If we have already proved the theorem and property mentioned, then there is nothing more to do. We can be certain that our answer is correct.

If we are just guessing, then we still need to prove that our answer is correct.

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

Proof

Let $\epsilon > 0$ be given. Choose $\delta = \epsilon$. (Note that $\delta > 0$.)

For any $x$ such that $0 < \left\mid x - 2 \right\mid < \delta$, we have

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

$= \left\mid \left(x + 3\right) - 5 \right\mid$ $\text{ }$ (Since $0 < \left\mid x - 2 \right\mid$, we know that $x \ne 2$, so we can simplify the fraction.)

$= \left\mid x - 2 \right\mid < \delta = \epsilon$.

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