1. ## limits/proof

A) find a q > 0 so that |x-3| < q implies that |x^2 +2x - 15| < 1/4

B) Prove this limit:
Lim x->5(x^2-3x+1) = 11

Lim x->0 |x|

2. Originally Posted by luckyc1423
Lim x->0 |x|
Limit is zero.

We need to find a interval I\{0} containting where I is open and containd 0.

Choose I=(-1,1)\{0}

If x_n is sequence in x converging to 0.
Then,
f(x_n)=|x_n| converges to |0|=0.

3. Originally Posted by luckyc1423
A) find a q > 0 so that |x-3| < q implies that |x^2 +2x - 15| < 1/4
if |x-3|< q then -q < x - 3 < q
so 3 - q < x < 3 + q

now we want |x^2 +2x - 15| < 1/4

since 3 - q < x < 3 + q, x^2 +2x - 15 is at least a little bigger than (3 - q)^2 +2(3 - q) - 15 = 9 -6q + q^2 + 6 - 2q - 15 = q^2 - 8q

so we want q^2 - 8q < 1/4
so 4q^2 - 32q - 1 < 0

=> q = [32 +/- sqrt(1024+16)]/8 = [32 + sqrt(1040)]/8 ~= 8.0311... since q > 0

also, since 3 - q < x < 3 + q, x^2 +2x - 15 is at most a little smaller than (3 + q)^2 +2(3 + q) - 15 = 9 + 6q + q^2 + 6 + 2q - 15 = q^2 + 8q

so we want q^2 + 8q < 1/4
=> 4q^2 + 32q - 1 < 0
=> q = [-32 +/- sqrt(1024 + 16)]/8 = (-32 + sqrt(1040))/8 ~= 0.0311... since q > 0

so q can be any value such that 0.0311 < q < 8.0311

so i overkilled this question, we could have stopped when i found the 8.0311 and just pick that, or pick 8 to be safe

4. Originally Posted by luckyc1423
B) Prove this limit:
Lim x->5(x^2-3x+1) = 11
I will not right out the formal proof.
I will just write out how to approach to it.

We can do this either with sequences or with d-e proof.
I assume you want a d-e proof.

5. Originally Posted by luckyc1423
B) Prove this limit:
Lim x->5(x^2-3x+1) = 11
so because they used the word "prove," i am hesitant to tell you to just plug in 5 for x in the function and show that you get 11 -- which you can do since the function is a polynomial and hence is continuous everywhere.

however, a proof is as follows.

Definition:
A limit is formally defined as follows: Let be a function defined on an open interval containing (except possibly at ) and let be a real number.
means that
for each real there exists a real such that for all where , .So, to prove Lim x->5(x^2-3x+1) = 11, we must show that for every e > 0 there exists a d > 0 such that for all x in dom(f) and 0 < |x - 5|< d implies |x^2-3x+1 - 11|=|x^2-3x-10|< e

Proof:

Let e > 0. choose d = e/|x+2|

then x in dom(f) and 0<|x-5|< d implies |x^2-3x+1 - 11|=|x^2-3x-10|=|(x - 5)(x + 2)|<|x - 5||x + 2|<e/|x + 2|*|x + 2| = e

therefore, the Lim x->5(x^2-3x+1) = 11

QED

You may want to choose the d that TPH suggested, since i don't think d can depend on x. at least you have the outline of the proof now