Question

Does anyone know how to 'prove limits' in pre-calculus? I have a test tomorrow and am plenty lost.

Answer 1

having the same problem lol

Answer 2

OMGOSH WHYYY....

Answer 3

cool, we can go over this.... either of you with a specific problem you'd like to look at?

Answer 4

ummmmmmm

Answer 5

hold on lemme find one

Answer 6

well what about the whole y=x/(x^2-2x)

Answer 7

as x-> ?

Answer 8

Find the limit if it exists, and explain why.
lim (2x+|x-3|)
x->3

Answer 9

@dfresenius we'll see if ariellllllll get's in to clarify here prob. statement... if not we'll do yours

Answer 10

gets*

Answer 11

nope. im lost.

Answer 12

are limits the same as asymptotes?

Answer 13

@Algebraic! gotcha

Answer 14

k doing @dfresenius prob.

Answer 15

okie doke

Answer 16

looking at the equation, do you see any continuity issues ? ie is there any sort of undefined value in that expression @dfresenius ?

Answer 17

hmm, the x is undefined?

Answer 18

there's not... there's an issue with differentiability at x=3 but there's no holes or infinities... ie 3-3 is zero (not a problem) ; 2*3 -0 is 6 (not a problem)

Answer 19

can you tell if it is discontinuous without graphing it?

Answer 20

yes @Arielllllllllllllllll look for zeros in the denom.

Answer 21

they indicate a 'problem'

Answer 22

ok, so a general strategy for solving this would be to plug in the limit?

Answer 23

OOOOHHHH. okaeeee.

Answer 24

your expression for example : x/(x^2-2x)
problem when x= 2 and problem when x=0 right?

Answer 25

yup.

Answer 26

they turn out to be two different types of problem, so both are instructive

Answer 27

what happens at x=2?

Answer 28

it becomes 0

Answer 29

numerator = 2 denominator = 0 so... we have an infinity...

Answer 30

2/0

Answer 31

moreover look what happens when you're approaching x=2 from the lefty side... what sort of values does the function have? try x=1 and x=1.99999 (use your calculator)

Answer 32

then try some numbers to the right of x=2 like x=3 and x=2.00001

Answer 33

is this helping either of you?

Answer 34

yes it is. thanks. and i think im getting it finally.

Answer 35

@dfresenius always try that strat first... if you get an infinity then you know the limit is undefined ... BUT if you get 0/0 it's a different kind of problem... which we'll talk about next..

Answer 36

so back to @Arielllllllllllllllll 's problem... we looked at the limit as x-> 2 for x/(x^2-2x)

Answer 37

so if it approacheds 2 from the left, y goes to -infinity?

Answer 38

it goes to -infinity if you approach x=2 from the left and it goes to +infinity if you approach x=2 from the right... what about at x=0 ??

Answer 39

yes @Arielllllllllllllllll that's right

Answer 40

omguhhhhh. it doesn't go anywhere?

Answer 41

it's the thing we mentioned 0/0 it would be tempting to call it an infinity but it's not..

Answer 42

when you see this, you know something is up....

Answer 43

so it's just 'undefined' or something like that?

Answer 44

it's time to break out your algebra and get to work...

Answer 45

ok, my book is solving it from both sides...
for 3+(positive side) they wrote (2x+x-3) = 3x-3 = 9-3 = 6
for 3-(negative side) (2x +3-x) = x+3 = 3+3 = 6
just curious if this is a good way of solving this

Answer 46

yes it is @dfresenius

Answer 47

ok, how will I know to only solve it from one side?

Answer 48

when x-> 0 x/(x^2-2*x) goes to 0/0 so... we need to re-arrange or simplify the expression to see what is ACTUALLY going on at that point..

Answer 49

scratch my last question

Answer 50

im getting it now

Answer 51

in this case it's pretty easy... there's an x in every term so it simplifies to 1/(x-2)

Answer 52

magic... we can now look at what happens as x-> 0 with no difficulty
1/(0-2) = -1/2

Answer 53

oh dang. then why am i in this stupid class? D; okay, okay, suppose it's only x/(x-2). The limit is at 2, and as you approach it from the left it goes to -infinity and from the right it goes to infinity?

Answer 54

@Arielllllllllllllllll yes that is correct:)

Answer 55

@dfresenius don't be afraid to look at it from both sides... in fact that's exactly what talking about the limit IS.... we are asking if the function approaches a well-defined value as x approaches some value from both sides..

Answer 56

if the function does approach a well defined value as x approaches some value from both sides... then we have a limit... if not, (or if it approaches infinity) then we don't have a limit :(

Answer 57

lim 2x-1/|2x^3-x^2|
x->0.5

Answer 58

x->.5-

Answer 59

this absolute value is really messing me up

Answer 60

good one. so... what's the denom at x=.5?

Answer 61

|.25-.25| = |0| = 0

Answer 62

yep

Answer 63

and the num. ? also zero... so we have one of those 'holes' 0/0

Answer 64

ok, so this means it ill only work from one side?

Answer 65

usually means a hole... in this case it actually is a bit more tricky... here's what we'll have to do I think... consider x>.5 what does the denominator do? ie is it positive or negative for x>.5 ?

Answer 66

i did x=1 and it ends up 1/1

Answer 67

it's positive... right? so for x>.5 the absolute value sign is meaningless... it does nothing...

Answer 68

yep

Answer 69

so 'ignore' it for x>.5 :

(2x-1)/(2*x^3-x^2) = (2x-1)/(x^2 *(2x-1) )

Answer 70

the " (2x-1) 's " cancel ...: 1/x^2

Answer 71

so you're looking at a limit of '4' as x->.5 from the right... that clear?

Answer 72

yep

Answer 73

cool... now the other side
when x<.5 what does the denom. look like?

Answer 74

then coming from the left is where we change the absolute value?

Answer 75

i used -1 and got -2

Answer 76

or -2/1

Answer 77

or wait, -3/1

Answer 78

damn my bad, its -3/-1 or just 3

Answer 79

yep... the denom. is negative for x<.5 (and x>0, but that's not a worry right now) the absolute makes it positive though so if we want to write it without the absolute value sign : (2x-1) / (-2x^3 +x^2) and we get (after algebra)
1/-x^2

Answer 80

so as x-> .5 - we get the limit is -4 ;)

Answer 81

that clear?

Answer 82

yep thank you so much for being patient!

Answer 83

No problem? all good? @Arielllllllllllllllll , if you're still there... any questions?

Answer 84

Im all set, thanks again!

Answer 85

Naw, I think I'm good. I needed a review, and this was plenty. Thanks :D

Answer 86

sure! @dfresenius

Answer 87

@Arielllllllllllllllll ok no problem! gl!