Question

Properties of Limits!
Medal & Fan! <3

Answer 1

http://www.saylor.org/site/wp-content/uploads/2011/11/2-3LimitProperties.pdf

Answer 2

I have 3 questions like this on my school work. And I can't figure out how to solve them.

Answer 3

Are you willing to answer my questions to understand what is going on ?

Answer 4

sure

Answer 5

If \(f(\color{red}{x})=\color{red}{x}\), then \(f(\color{blue}{y})= ??\)

Answer 6

y

Answer 7

woah...yes
\(f(z) =?\)

Answer 8

z

Answer 9

yes, f(2) =?

Answer 10

2

Answer 11

yup, simply replace x by 2, right?

Answer 12

yes

Answer 13

Now, if f(x) = x^2+x+1, and I want f(2), what should I do?

Answer 14

plug in 2 where i see x?

so f(2) = 2^2+2+1

Answer 15

Bingo!! simplify, it is f(2) =7 right?

Answer 16

yes... but i was confused by the x & h...
is x and h the same in this case?

Answer 17

be patient!! we get it now.

Answer 18

now, I want f(2+h), what should I do?
Recall, f(2) , replace x by 2

Answer 19

set them equal to each other??
like..
2+h = 2

Answer 20

nope, I do it again for you to see what we should do
\(f(\color{red}{x})= \color{red}{x}^2 + \color{red}{x}+1\)
\(f(\color{blue}{2})= \color{blue}{2}^2 + \color{blue}{2}+1=7\)
\(f(\color{green}{2+h)})= \color{green}{(2+h)}^2 + \color{green}{(2+h)}+1\) ok???

Answer 21

ohhh ok!

Answer 22

Now, you open the parentheses for me, what do you have for f(2+h)?

Answer 23

7+h^2+h

Answer 24

3h or h?? I meant 7 + h^2 +h or 7 + h^2 +3h??

Answer 25

where has the 3 come from?

Answer 26

\((2+h)^2 +(2+h)+1= 2^2 +2h +h^2 +2+h+1= 7 + 3h +h^2\)

Answer 27

Remember?? \((a+b)^2 = a^2+2ab +b^2\)

Answer 28

ohh ok

Answer 29

Combine f(2+h) -f(2)

Answer 30

\(f(2+h) = 7+3h+h^2\\f(2) =7\\--------------------\\=3h+h^2\)
ok?

Answer 31

alright

Answer 32

factor h out,

Answer 33

you have
\[\lim_{h\rightarrow 0}\dfrac{f(2+h)-f(2)}{h}=\lim_{h\rightarrow 0}\dfrac{h(3+h)}{h}\]

Answer 34

cancel h , you have
\[\lim_{h\rightarrow 0}3+h\]
Ok? as \(h\rightarrow 0\) , the limit goes to ????

Answer 35

3!!

Answer 36

Woah... you got it fast.

Answer 37

:)

Answer 38

I may need your help for another.. if that is ok?

Answer 39

I hope it is not the same with this problem. Because you showed me you got it, hence there is no reason for me to help you with the same kind of this. Am I right?

Answer 40

Stick around just in case I need help? just for a few more minutes.
Thank you very much for your help

Answer 41

I figured out all 3 questions. thank you for your help