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

Oct 11, 2015

See the explanation.

#### Explanation:

I assume that the statement we want to prove is ${\lim}_{x \rightarrow - 3} \left({x}^{2} + 3 x\right) = 0$

Preliminary Work

The $\epsilon , \delta$ proofs for quadratic functions are more involved than those for linear functions. Most of us need to be shown this method, we don't come up with it by ourselves.

We begin (as we do for linear functions) by examining the difference between $f \left(x\right)$ and $L$.
Keep in mind that, for his question #abs(x-a) = abs(x-(-3)) = abs(x+3)

Now,
$\left\mid \left({x}^{2} + 3 x\right) - 0 \right\mid = \left\mid x \left(x + 3\right) \right\mid = \left\mid x \right\mid \left\mid x + 3 \right\mid$

We can control, through the choice of $\delta$, the size of $\left\mid x + 3 \right\mid$.

If $\left\mid x + 3 \right\mid$ is small, then $x$ is close to $- 3$. We'll put a bound on $\left\mid x \right\mid$ by choosing $\delta$ so that is is less than some positive number we select now.
$1$ is easy to work with, so we'll make sure that $\delta \le 1$.
This will assure us that
when $\left\mid x - \left(- 3\right) \right\mid < \delta \le 1$, we have

$x$ is within $1$ unit of $- 3$,

So, $- 3 - 1 < x < - 3 + 1$

I.e. $- 4 < x , - 2$,

so $\left\mid x \right\mid < 4$.

If we also make sure that $\delta \le \frac{\epsilon}{4}$, the we will have:

When $\left\mid x + 3 \right\mid < \delta$, then

$\left\mid x \right\mid \left\mid x + 3 \right\mid < 4 \left\mid x + 3 \right\mid < 4 \left(\frac{\epsilon}{4}\right) = \epsilon$

Proof

Given $\epsilon > 0$, choose $\delta = \min \left\{1 , \frac{\epsilon}{4}\right\}$

If $0 < \left\mid x + 3 \right\mid < \delta$, then

$\left\mid \left({x}^{2} + 3 x\right) - 0 \right\mid = \left\mid x \right\mid \left\mid x + 3 \right\mid < 4 \left\mid x + 3 \right\mid \le 4 \left(\frac{\epsilon}{4}\right) = \epsilon$

That is, $\left\mid \left({x}^{2} + 3 x\right) - 0 \right\mid < \epsilon$.