Question

Using the formal definition of limits, prove that lim(x -> -1.5) (9-4x^2)/(2x+3) = 6.

So far I have:
|((-2x+3)(2x+3))/(2x+3)-6| < e
|-2x -3| < e

But I don't know if I should do:
|x+3/2|> -e/2
or to just leave it. Someone please help!

Answer 1

That last step is wrong, because the negative never makes it outside the absolute value sign.

Write f(x) = (9-4x^2)/(2x+3). Then what you have have is
| f(x) - 6 |
= | (-2x+3)(2x+3)/(2x+3) - 6 |
= | -2x + 3 - 6 |
= | 2x + 3 |
= 2 | x - (-3/2) |

Now let epsilon e be an arbitrary number e > 0. Then choose delta d = e/2. Then
0 < | x - (-3/2) | < d = e/2 ==> | f(x) - 6 | < e

Therefore by definition of the limit:
\[\lim_{x \rightarrow -3/2} \ \ f(x) = 6.\]

Answer 2

Epsilon-delta proofs are probably the most subtle mathematics you've learnt so far.

So do yourself a favor (and I'll feel better about it too!). Take a blank piece of paper and write out the proof above and make sure you understand every step, particularly at the end.

When you can do that with comfort without looking at what we're written here, then you've really learnt something.

Answer 3

Thanks James! I appreciate your help. I actually ended up figuring it out. I tend to make minor mistakes with my math, especially when I am tired. I understand it now though! :D

Answer 4

Hey James, can you help me again?
I don't know if I did the following question right:
"Prove that \[\lim_{x \rightarrow -\infty} (1 + x^3) = -\infty\] using the precise definition of infinite limits (i.e. x < N then f(x) < M)."

I did:

1 + x^3 < M
x < (M -1)^(1/3)

N > 0, so N = max(1, (M - 1)^(1/3))
x < N = max(1, (M - 1)^(1/3))
1 + x^3 < max(1+x^3,M) < M

Therefore, \[\lim_{x \rightarrow -\infty} (1 + x^3) = -\infty\]

Answer 5

It is not the case that N > 0 and
"1 + x^3 < max(1+x^3,M) < M"
this step isn't quite right.

Go through it again carefully.

Answer 6

I am confused :|

Answer 7

Go back to the definition. The idea here is that for all M < 0, we can find an N such that
\[x < N \implies f(x) < M\]

So where exactly is the clean logic here. Show me how to start with an M and find such an N that gives us f(x) < M.

You've almost got it, but you have a couple of non sequitors in your logic, such as the two I wrote above.

Answer 8

Oh, I thought M and N were supposed to be positive, but they are supposed to be negative?

Answer 9

clearly because we're taking the limit as x --> NEGATIVE infinity and the limit we want is also negative infinity.

Answer 10

So we want to get f(x) arbitrarily large and negative provided x is sufficiently large and negative.

Answer 11

Okay, so do I have up to x < (M-1)^(1/3)) correct?

Answer 12

Yes, that's all fine.

Answer 13

and that's the preamble you want to knowing how to choose N given M

Answer 14

Okay
So then N < 0.. so do we still take max(1, (M-1)^(1/3))?

I know I am lost here, but my teacher hasn't shown us how to do this and I watched a video on x approaching positive infinity, and I got confused.

Answer 15

Hence ...

given M < 0, choose N = (something involving M). Then

x < N ==> f(x) < M

and therefore by definition the limit as x --> -infty of f(x) = -infty.

Answer 16

So given M < 0, what is your choice for N which will give us the logical deduction we want?

Answer 17

I'm not sure what you are asking

Answer 18

It's exactly like in the epsilon-delta proof. Given epsilon, what choice of delta will give us
0 < |x-a| < delta ==> | f(x) - L | < epsilon

Answer 19

Okay... Could you walk me through this?
We aren't going to be taught in class because we are behind, and I am obviously not getting it.

Answer 20

Here, the formulation of the limit is this:
if for all M < 0 (i.e., large and negative) we can find an N such that

x<N ==> f(x)<M

Then we will say that is the definition that
\[\lim_{x \rightarrow -\infty} \ f(x) = -\infty\]

Answer 21

Okay, I get that much

Answer 22

Ok, so given M , what choice of N should we make such that

x<N ==> f(x)<M

Answer 23

Given M < 0, let N = some expression involving M, then

x<N ==> f(x)<M

Answer 24

What is that expression for N?

Answer 25

Is that what I solved for? As in (M - 1)^(1/3)?

Answer 26

Yes. Given M < 0, let N = (M-1)^(1/3), then

x < N ==> x < (M-1)^(1/3)
==> 1 + x^3 < M

Hence by definition ....

Answer 27

Oh! That makes sense
So, by definition the limit is -infinity... if that is what you were asking?

Answer 28

No, that's what YOU are asking. And yes.

Answer 29

What I don't blame you for getting confused about, is bounding these numbers to be large, negative.

Answer 30

Thanks so much! I know I can be a bother. I just feel lost at times when my teacher jumps over things. You are a great help!

Answer 31

But work this example again and make up your own example for the other three types of limits x goes to plus/minus infinity and the limit is plus/minus infinity.

Answer 32

Figure out exactly what are the definitions, and then make up some of your own examples and work them through.

Answer 33

I will :)

Answer 34

Now, I think I deserve a medal.

Answer 35

Of course
I forget about those

Answer 36

James, could you help me with the following question? Please!

Show that

det(A) =1/2 |tr(A) 1|
|tr(A^2) tr(A)|

for every 2 x 2 matrix.

Answer 37

Doing some work of my own right now ... give me a few

Answer 38

No problem :)

Answer 39

OK, so writing A =
a b
c d

then A^2 =
a^2 + bc ab + bd
ac + cd bc + d^2

Hence tr(A^2) = a^2 + 2bc + d^2

Answer 40

and tr(A) = a + d of course

Now the determinant of that matrix is tr(A)^2 - tr(A^2)
= (a+d)^2 - (a^2 + 2bc + d^2)
= 2 (ad - bc)
= 2 det(A)

Answer 41

So this is straightforward, certainly much easier than your limit problems.

Answer 42

And frankly, I'd bet a dollar this is a completely useless identity. I think the question was just made up to get your comfortable manipulating these expressions from matrices

Answer 43

I believe so :)
Thank you very much! You are a great help :)

Answer 44

Hey James. I have another quick problem that I am a little confused about. Our prof wanted us to attempt it before our lesson, but that is not going well. I was wondering if you could help me with it?

Consider the vector v = (1,-5)
a) Find all values of k such that u = (k,7) is orthogonal to v
b) Find all values of k such that u = (k,7) is parallel to v

For a, I computed the dot product because I know that when two vectors are perpendicular they have a dot product of zero. So I found that when k = 35, u is orthogonal to v. I am not sure if that finds all values of k, or if I chose the right method. But I think it works.

I am not sure how to attempt b. I know that two parallel vectors are scalar multiples of each other, but I am little confused.

Answer 45

So for a, you're exactly right:
the two vectors are orthogonal if and only if
(k,7).(1,-5) = 0 <=> k - 35 = 0 <=> k = 35
and hence k = 35 is the only value of k for which they are orrthongal.

Answer 46

Now (k,7) is parallel to (1,-5) if there exists a nonzero scalar, c such that
(k,7) = c(1,-5) <=> (k,7) = (c,-5c)

Now that is the case if both
k = c
and
7 = -5c

By the second of those equation c = -7/5 and now by the first k = -7/5 also.

Answer 47

Oh, so u is parallel to v if k = -7/5? That makes sense. I remember doing something like that in grade 12.

Thanks for your help!

Answer 48

sure

Answer 49

Hey James, when you have a moment, would you mind looking at my other question? Please and thank you!