Question

Limits again.

Answer 1

Thank You again, I preciate your help and contribution to my learning!!!

Answer 2

Alright. Let's talk about some properties of limits. It's going to be similar to properties of exponents or properties of multiplication.

Answer 3

Suppose we have a constant function. For example \(f(x)=5\). No matter what \(x\) is, then \(f(x)\) always is equal to \(5\).

Answer 4

It shouldn't be shocking to realize that \[
\lim_{x\to a}f(x)=5
\]This is regardless of \(a\). Even if \(a=2\) or \(a=99\) it will always happen.

Answer 5

So it would not matter what x is equal to at all, and whether it is equal to a 100 or 3 the graph will be the same?

Answer 6

Yes. \(f(x)\) will always approach \(5\).

Answer 7

And the difference b/w x=4 and x=100 when f(x)=5,
is just going to be "how close is it to 5", right?

Answer 8

So we write this property as: \[
\lim_{x\to a}c=c
\]And we say \(c\) is a constant with respect to \(x\).

Answer 9

Well, in all cases it will be equal to \(5\). It's never going to be not equal to 5 or a little bit under 5, for example.

Answer 10

I see!

Answer 11

The graph is:
In these cases if you let \(a=-10\) or \(a=10\) then you see it is still \(5\).

Answer 12

10 and -10, where are they from?

Answer 13

Okay, the next property of limits:
Suppose we know the limit \(f(x)\to L\) and \(g(x)\to K\) when \(x\to a\).
Then we know that \([f(x)+g(x)]\to [L+K]\) when \(x\to a\).

Answer 14

\(10\) and \(-10\) were randomly selected.

Answer 15

Does that make sense?

Do you understand what I mean about the adding thing?

Answer 16

Your comment when you say "Okay, the next property of limits...."
I got lost there!

Answer 17

I don't get the formulas f(x) arrow L
g(x) arrow K
what is the difference b/w those two?

Answer 18

Okay, so remember \[
g(x) = \frac{(x-2)(x+1)}{x-2}
\]We said\[
\lim_{x\to 2}\frac{(x-2)(x+1)}{x-2}=3
\]

Answer 19

So g(x) is a regular function and with L it is limited to a certain value, correct?

Answer 20

And for \[
f(x) =5
\]We said \[
\lim_{x\to a}5=5
\]This even works when \(a=2\) \[
\lim_{x\to 2}5=5

\]

Answer 21

What the addition property says is: \[
\lim_{x\to 2}5+\frac{(x-2)(x+1)}{x-2} =5+3 = 8
\]

Answer 22

and the last formula works and is about/approximately the same for any value of a?

Answer 23

Yeah, that formula for constants works for any value of \(a\).

Answer 24

However, this addition property only works if \(a\) is the same value in both cases.

Answer 25

So \(a\) had to be \(2\) in both cases for it to work.

Answer 26

Now, I confused you with the arrows, right? Let me explain it a bit.

Answer 27

When I write the \(f(x)\to L\) I'm almost saying something like \(f(x)\approx L\). The point is that \(f(x)\) gets closer and closer to \(0\).

Answer 28

get it! Thank A lot man!!!!!

Answer 29

I get the arrow!

Answer 30

I mean \(L\) not \(0\). That was a typo. \(f(x)\) gets closer to \(L\).

Answer 31

Yep! In fact you could use arrows to show a limit, but the \(\lim\) notation is more popular and familiar.

Answer 32

Yeah, I didn't see the zero, when i saw a symbol for approximately equals I already got it...

Answer 33

Okay, do you understand the addition property of limits?

Answer 34

Yes! I think I do!

@wio, can we go to a different question soon inkyvoyd constantly distracts me by typing and not posting, b/c it keeps saying that he is typing.

Answer 35

(and you will get more medals)

Answer 36

message me!

Answer 37

Okay one moment.

Answer 38

Thank You!