Question

Question #4c5ac

Answer 1

See below.

Explanation:

Prove: `lim_(x->-3)x^2+4x+1=-2`

Work (not part of proof):

`0 < |x+3| < delta`, `|(x^2+4x+1)+2| < epsilon`

We need to manipulate the `|(x^2+4x+1)+2| < epsilon` to be `|x+3|<"something"` to set delta equal to that term:
`|(x^2+4x+1)+2| < epsilon`

`|x^2+4x+3| < epsilon`

`|(x+3)(x+1)| < epsilon`

`|x+3| < epsilon/|x+1|`

Since we cannot have an `x` term with epsilon, we let `delta=1` and solve for the value of `x+1`:
`0 < |x+3| < 1`

`-1 < x+3 < 1`

`-3 < x+1 < -1`

Here, we choose the larger value since if we chose the smaller value, the `-3` would not be included, so:
`|x+1|<3`

Therefore,
`|x+3| < epsilon/3`

Proof:

`forall` `epsilon > 0`, `exists` `delta > 0` such that:
if `0 < |x+3| < delta`, then `|(x^2+4x+1)+2| < epsilon`.
Given `0 < |x+3| < delta`, let `epsilon = min(1,epsilon/3)`:

`0 < |x+3| < epsilon/3`

`0 < 3|x+3| < epsilon`

`0 < |x+1||x+3| < epsilon`

`0 < |x^2+4x+3| < epsilon`

`0 < |(x^2+4x+1)+2| < epsilon`

`therefore` `lim_(x->-3)x^2+4x+1=-2`

Answer 2

Since `f(x) = x^2+4x+1` is a polynomial, it is continuous for every `x in RR`, then:

`lim_(x->-3) f(x) = (-3)^2+4*(-3)+1 = 9-12+1 = -2`

We want to prove that given `epsilon >0` we can find a corresponding `delta_epsilon` such that:

`abs(x-(-3)) < delta_epsilon => abs (x^2+4x+1-(-2)) < epsilon`

that is:

`abs(x+3) < delta_epsilon => abs (x^2+4x+3) < epsilon`

Evaluate:

`abs (x^2+4x+3) = abs((x+3)(x+1)) = abs(x+3)abs(x+1)`

Now for `abs(x+3) < delta_epsilon` we have:

`abs(x+1) = abs(x+3-2)`

and based on the triangular inequality:

`abs(x+1) <= abs(x+3)+abs(2)`

that is:

`abs(x+1) < delta_epsilon+2`

Given an arbitrary number `epsilon > 0` we can choose then `delta_epsilon < min(1, epsilon/3)` and we have that because `delta_epsilon < 1`:

`abs (x+1) < 2+delta_epsilon < 2+1 = 3`

and because `delta_epsilon < epsilon/3`

`abs(x+3) < delta_epsilon < epsilon/3`

but then:

`abs(x+3)abs(x+1) < 3 xx epsilon/3 = epsilon`

and in conclusion:

`abs (x^2+4x+3) < epsilon`

which proves the point.