# How do you use the epsilon delta definition to find the limit of ((9-4x^2)/(3+2x)) as x approaches -1.5?

Apr 20, 2016

See the explanation section below.

#### Explanation:

The definition of limit is not really useful for finding limits.
It is used to prove that the limit is what I said it is.

To find this limit

If we try to find ${\lim}_{x \rightarrow - \frac{3}{2}} \frac{9 - 4 {x}^{2}}{3 + 2 x}$ by substitution we get the indeterminate form $\frac{0}{0}$. We need some technique to find the limit.

Both the numerator and denominator are polynomials and they share a zero. That tells us that they share a factor, so the expression can be simplified.

$\left(9 - 4 {x}^{2}\right)$ is a difference of squares, so we factor and reduce:

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

$= 3 - 2 x$ $\text{ }$ provided that $x \ne - \frac{3}{2}$

Since the limit doesn't care what happens when $x$ is equal to $- \frac{3}{2}$, but only when $x$ is close to $- \frac{3}{2}$, we can finish:

lim_(xrarr-3/2)(9-4x^2)/(3+2x) = lim_(xrarr-3/2)(3-2x) = 3-2(-3/2) = 6

Proving that the limit is 6

Claim: ${\lim}_{x \rightarrow - \frac{3}{2}} \frac{9 - 4 {x}^{2}}{3 + 2 x} = 6$

Proof:

Let $\epsilon > 0$ be given. Choose $\delta = \frac{\epsilon}{2}$

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

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

$= \left\mid \left(3 - 2 x\right) - 6 \right\mid$ $\text{ }$

(Observe that $0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid$ implies $x \ne - \frac{3}{2}$. So, it's OK to reduce.)

$= \left\mid \left(3 - 2 x\right) - 6 \right\mid = \left\mid - 2 x - 3 \right\mid$

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

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

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

$< 2 \delta$

$= 2 \left(\frac{\epsilon}{2}\right) = \epsilon$

We have shown that, for any positive $\epsilon$, there is a positive $\delta$ such that: for all $x$,

$0 < \left\mid x - \left(- \frac{3}{2}\right) \right\mid < \delta$ implies $\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid < \epsilon$

Therefore, by the definition of limit, ${\lim}_{x \rightarrow - \frac{3}{2}} \frac{9 - 4 {x}^{2}}{3 + 2 x} = 6$.