Question

For Opcode

Answer 1

@Opcode Yo come here

Answer 2

Do you understand for example, how to evaluate: \[
\lim_{x\to 1}x^2+1
\]?

Answer 3

Not really, I just know:
\[\lim_{x\to 5}x-4=1\]
So not really with the exponents...

Answer 4

Do you know what a polynomial is?

Answer 5

Yes.

Answer 6

Well, for any polynomial, the limit is just the same as plugging in the number.

Answer 7

Here is a tougher one: \[
\lim_{x\to 2}\frac{x^2-4}{x-2}
\]

Answer 8

You have to use difference of squares to factor the numerator: \[
a^2-b^2 = (a-b)(a+b)
\]

Answer 9

Is the number four the answer?

Answer 10

Yeah...

Answer 11

How'd you get that?

Answer 12

Hmm.
No, it's 0...
Wait, 4-4=0
2-2=0
>.< I'm confused, I put four because you said difference. I subtracted..

Answer 13

Well, what does \(0/0=\)?

Answer 14

It's undefined?

Answer 15

Are you telling me that \(0/0=4\)?

Answer 16

No...

Answer 17

Actually, \[
\frac{0}{0}
\]Is not undefined...

Answer 18

Think about it this way, we want a number \(x\) such that... \[
x\times 0 = 0
\]Are you telling me no such number is defined?

Answer 19

Cause I'd say \(x=4\) is a number that satisfies the equation
=]

Answer 20

It's not? But I thought you can't divide by 0...
Oh wait, I see! I get it now.

Answer 21

Give me another one. I want to make sure I'm doing it right. Please :3

Answer 22

Well, hold on a second... first of all.... consider \[
\frac{1}{0}
\]This is undefined because the question is \(x\) such that \[
x\times 0 = 1
\]

Answer 23

Undefined just means we know that our current sets of numbers can not solve the equation.

Answer 24

At some point \(\sqrt{-1}\) was considered undefined for the same reason. We came up with \(i=\sqrt{-1}\) (the imaginary and complex number set) because it was beneficial. We never came up with something like \(j=1/0\), because it has no known benefit.

Answer 25

I see okay.
\[\lim_{x\to 4}\frac{x^2-4}{x-2}\]
16 - 4 = 12
4 - 2 = 2
\[\frac{12}{2} = 6\]
\[x \times 2 = 12\]
\[4 \times 2 = 12\]
So I'm correct?

Answer 26

In the case of \[
\frac{0}{0}
\]It's not that there is no number which will satisfy the question. The issue is that too many numbers satisfy the question. We can't determine what the answer is without some context. That is we call it an indeterminate form.

Answer 27

Hmmm? What you did was wrong. You can't just change the point your approaching in the limit from 2 to 4. I'll show you how to do it.

Answer 28

But I just want you to think a bit about what I said involving "undefined" vs "indeterminate form"

Answer 29

Ah, okay. I'll remember undefined and indeterminate form.

Answer 30

The solution is as follows: \[\begin{split}
\lim_{x\to 2}\frac{x^2-4}{x-2}
&= \lim_{x\to 2}\frac{(x-2)(x+2)}{x-2} \\
&= \lim_{x\to 2}x+2 \\
&= 2+2\\
&=4\end{split}
\]

Answer 31

The \(x-2\) cancels when I factor.

Answer 32

Wow, I was doing it wrong :O
But now I understand. Thanks! ^_^.

Answer 33

Yeah, but one thing to consider... suppose we had two functions:\[
f(x) = \frac{x^2-4}{x-2} \\
g(x) = x+2
\]As it turns out, these functions are always equal.

Answer 34

Notice that \(f(2)\) is undefined while \(g(2)=4\)

Answer 35

We could say: \[
f(x) = \begin{cases}
g(x) &x\ne 2\\
\text{undefined} & x=2
\end{cases}
\]

Answer 36

However... the following is always true: \[
\lim_{x\to a} f(x) = \lim_{x\to a}g(x)
\]Even when \(a=2\) it is true.

Answer 37

@Opcode Does that make sense?

Answer 38

Yes, somewhat. But why is it always equal? I don't understand that part...

Answer 39

Well, notice how the only place where they aren't equal is at 2?

Answer 40

The only reason they aren't equal at 2 is because \(f(x)\) runs into a division by 0 problem.

Answer 41

Okay, I understand now. \(f(x)\) is undefined because of the division by zero.

Answer 42

The limits allow us to get around division by zero problems, sometimes.

Answer 43

Mainly when we have indeterminate forms, limits allow us to find out about what a function 'should be doing' but can't, due to something like division by zero.

They also let us know what a function should do when it tends to infinity, which is really helpful.

Answer 44

This is sort of what I meant by 'should be doing'
That line isn't supposed to represent \(f\) or \(g\) by the waa, it's just a random line.

Answer 45

The point it, limits allow us to 'fill in the gaps' in a way.

Answer 46

In my opinion, limits are the coolest thing that came from Calculus. You need them to do derivatives and integrals.

Answer 47

Well now I know the purpose for them. At first I thought they were just useless to learn :O

Answer 48

Well, let's talk about derivatives just for a bit.