# How do you prove the statement lim as x approaches -1.5 for  ((9-4x^2)/(3+2x))=6 using the epsilon and delta definition?

Oct 18, 2015

You can't.

#### Explanation:

You can't prove it since it isn't 6.
${\lim}_{x \to - 1.5} \frac{9 - 4 x}{3 + 2}$ does not exist. The limit from the left is $+ \infty$ and the limit from the right is $- \infty$.

Oct 19, 2015

See the explanation.

#### Explanation:

Preliminary analysis
We need to show that for every positive $\epsilon$, there is a positive $\delta$ such that if $0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid < \delta$ we get $\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid < \epsilon$.

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

For every $x$ other than $- \frac{3}{2}$, we have:

$\frac{9 - 4 {x}^{2}}{3 + 2 x} = 3 - 2 x$.

So, for every $x$ other than $- \frac{3}{2}$, we have:

$\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid = \left\mid \left(3 - 2 x\right) - 6 \right\mid$

$= \left\mid - 2 x - 3 \right\mid = \left\mid \left(- 4\right) \left(x + \frac{3}{2}\right) \right\mid = \left\mid - 4 \right\mid \left\mid x + \frac{3}{2} \right\mid$.
$= 4 \left\mid x + \frac{3}{2} \right\mid$

We want this to be less than $\epsilon$ and we control, through $\delta$, the size of $\left\mid x + \frac{3}{2} \right\mid$

If we make $\left\mid x + \frac{3}{2} \right\mid < \frac{\epsilon}{4}$, then we will have

$4 \left\mid x + \frac{3}{2} \right\mid < 4 \left(\frac{\epsilon}{4}\right) = \epsilon$ as desired.

Now we are ready to write the proof:

Proof

Given $\epsilon > 0$, let $\delta = \frac{\epsilon}{4}$. Observe that this $\delta$ is also positive, as required.

Now if $x$ is chosen so that $0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid < \delta$, the we have:

$\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid = \left\mid \left(3 - 2 x\right) - 6 \right\mid$

$= \left\mid - 2 x - 3 \right\mid = \left\mid - 4 \right\mid \left\mid x + \frac{3}{2} \right\mid$.

$= 4 \left\mid x + \frac{3}{2} \right\mid < 4 \delta = 4 \left(\frac{\epsilon}{4}\right) = \epsilon$

That is: if $0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid < \delta$, then $\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid < \epsilon$.

So, by the definition of limit, ${\lim}_{x \rightarrow - 1.5} \left(\frac{9 - 4 {x}^{2}}{3 + 2 x}\right) = 6$