Question

lim h ->-2 (h^3 +8)/(h+2) can anyone help me?

Answer 1

hint:

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

Answer 2

Jim, can you elaborate any?

Answer 3

since 8 = 2^3

this means that for h^3 + 8 = h^3 + 2^3, we can see that

a = h
b = 2

Answer 4

therefore,

h^3 + 8

factors to

(h+2)(h^2 - 2h + 4)

using the first rule I posted

Answer 5

so

\[\large \frac{h^3 + 8}{h+2}\]

becomes

\[\large \frac{(h+2)(h^2 - 2h + 4)}{h+2}\]

Answer 6

so then I just replace H with -2 and simplify it?

Answer 7

simplify, then plug in h = -2

Answer 8

umm, I get 0/0

Answer 9

so is it DNE?

Answer 10

look at

\[\large \frac{(h+2)(h^2 - 2h + 4)}{h+2}\]

notice anything cancelling?

Answer 11

so it's h^2 - 2h + 4? so 4+4+4 so the limit is 12?

Answer 12

correct

Answer 13

Wonderful! one more question

Answer 14

alright

Answer 15

go for it

Answer 16

http://puu.sh/4InZm.jpg

Answer 17

limit is -6 correct?

Answer 18

hmm not sure about the double bar notation

Answer 19

looks like absolute value, but why did they use 2 pairs instead of 1?

Answer 20

greatest integer function, the thing when you round up

Answer 21

oh ok

Answer 22

like 2.01 = 3

Answer 23

i gotcha

Answer 24

aka, the ceiling function

Answer 25

or wait, is that the floor, this page says otherwise

http://www.mathwords.com/f/floor_function.htm

Answer 26

it's the way that graph looks

Answer 27

wait no it rounds up not down

Answer 28

hmm odd

Answer 29

that page says to round down

Answer 30

I think that's a different version

Answer 31

yeah it's a bit ambiguous

Answer 32

I prefer floor and ceiling notations

Answer 33

they are definitely more clear

Answer 34

Are you doing problems from a book? If so, what is the way the function is defined there?

Answer 35

also, is this: http://puu.sh/4IonL.png zero?

Answer 36

0 is correct

Answer 37

also, http://puu.sh/4Iovu.jpg is zero as well??

Answer 38

correct again

Answer 39

can you help me for this? http://puu.sh/4IoIm.jpg

Answer 40

hint: rationalize the numerator by multiplying top and bottom by \(\large \sqrt{x}+5\)

this is the conjugate of the numerator

Answer 41

is the answer 0 again?

Answer 42

no

Answer 43

what is the conjugate?

Answer 44

in general if you have a+b

the conjugate is a-b

Answer 45

when you multiply the two, you get a difference of squares

(a+b)(a-b) = a^2 - b^2

Answer 46

so the conjugate of \(\large \sqrt{x}-5\) is \(\large \sqrt{x}+5\)

Answer 47

multiply the two to get ????

Answer 48

x - 25?

Answer 49

good

Answer 50

now multiply the denominator by \(\large \sqrt{x}+5\)

this is so you keep things balanced

Answer 51

notice how the x-25 terms will cancel

Answer 52

this leaves you with

\[\large \frac{1}{\sqrt{x}+5}\]

Answer 53

is the limit 1/10?

Answer 54

yep, you got it

Answer 55

okay, so for this one http://puu.sh/4IpJO.jpg A and B are 0 and C is 2, correct?

Answer 56

this will be the last one I help you with today

Answer 57

that's fine :D

Answer 58

all 3 answers you got are incorrect unfortunately

Answer 59

try plugging in x = 3.01, then plug in x = 3.001

Answer 60

the graph looks like http://puu.sh/4IpJO.jpg correct?

Answer 61

what do you get as you get closer to x = 3?

Answer 62

there is no graph in that post

Answer 63

oops

Answer 64

http://puu.sh/4IpZQ.png

Answer 65

much better

Answer 66

so as you approach x = 3 from the right, where do you get close to (in terms of y)?

Answer 67

shoot I made a silly mistake. A and B are 1?

Answer 68

C is also 1 because the left and right hand limits are both 1

Answer 69

remember that the limit only exists if the left and right limits are equal

in this case, the limiting value equals the left and the right hand limits

Answer 70

Wonderful! thank you so much for your help, I think I understand it better now

Answer 71

that's great