Question

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

Answer 1

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_(xrarr-3/2)(9-4x^2)/(3+2x)` by substitution we get the indeterminate form `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.

`(9-4x^2)` is a difference of squares, so we factor and reduce:

`(9-4x^2)/(3+2x) = ((3+2x)(3-2x))/(3+2x)`

`= 3-2x` `" "` provided that `x != -3/2`

Since the limit doesn't care what happens when `x` is equal to `-3/2`, but only when `x` is close to `-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_(xrarr-3/2)(9-4x^2)/(3+2x) = 6`

Proof:

Let `epsilon > 0` be given. Choose `delta = epsilon/2`

Now if `x` is chosen so that `0 < abs(x-(-3/2)) < delta`, then we will have

`abs((9-4x^2)/(3+2x) - 6) = abs(((3+2x)(3-2x))/(3+2x)-6)`

`= abs((3-2x)-6)` `" "`

(Observe that `0 < abs(x-(-3/2))` implies `x != -3/2`. So, it's OK to reduce.)

`= abs((3-2x)-6) = abs(-2x-3)`

`= abs((-2)(x+3/2))`

`= abs(-2)abs(x+3/2)`

`= 2abs(x+3/2)`

`< 2delta`

`= 2(epsilon/2) = epsilon`

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

`0 < abs(x-(-3/2)) < delta` implies `abs((9-4x^2)/(3+2x) - 6) < epsilon`

Therefore, by the definition of limit, `lim_(xrarr-3/2)(9-4x^2)/(3+2x) = 6`.