Question

@jim_thompson5910

Clarification with some limits

Answer 1

Okay so, even though I know they're easy, I at times can't grasp limits approaching infinity/piecewise approaching infinity

Answer 2

And I'm talking about analytically, not from looking at a graph

Answer 3

ok go ahead and post the problem

Answer 4

Oh an also absolute values, so we'll start with that.

We plug in the limit and then use the absolute value right? Lol

So with something that was:

\[Lim x-> 5 \frac{ x }{ \left| x-5 \right| }\]
That would be DNE so I'd have to go from 5+ and 5- to see if it's positive or negative right?

Answer 5

if you're doing this analytically, I'd recommend setting up a table

So you'd approach x = 5 from the left by starting with x = 4.9, then moving to x = 4.95 then going to x = 4.99 etc etc

Answer 6

you'd plug each x value into that function to get the corresponding y value and see what happens as you get closer to x = 5

Answer 7

Yeah you plug in values to the left and right of what you're approaching right?

Answer 8

exactly

you calculate the left hand limit (LHL) and the right hand limit (RHL)

Answer 9

How do I know when to answer something with that the limit is approaching +- infinity?

Answer 10

Because when a limit doesn't exist, it's technically approaching +- infinity.

Answer 11

well you don't know for sure but as you get closer to the target value, the results just explode on you in terms of size

for instance,
when x = 4.9, y = 49
when x = 4.95, y = 99
when x = 4.99, y = 499
when x = 4.999, y = 4999
when x = 4.9999, y = 49999

so y is heading off to positive infinity as x approaches 5 from the left

Answer 12

Yeah so that would be my answer right?

Answer 13

same thing for the right hand limit, just with different x values

Answer 14

Well I mean my answer is DNE but the +- infinity proves it

Answer 15

yes, it's DNE assuming you don't consider infinity to be an answer for limits

sometimes I've seen some books just report +infinity or -infinity instead of DNE

Answer 16

honestly, I'd go with +infinity since that's what I'm used to

Answer 17

DNE seems to suggest that LHL = RHL is false

but the two one-sided limits produce +infinity

Answer 18

Yeah and it's if it's as simplified as you can get it, and it's still DNE that's when your answer is in terms of infinity right?

Answer 19

I'm not 100% sure, let me do some research on that subtlety

Answer 20

I got that from my notes, I was just double checking

Answer 21

this is also saying what you are saying

http://www.themathpage.com/acalc/infinity.htm

see attached

Answer 22

What if you had like... Lim x-> infinity of 6

It's just 6 right? lol

Answer 23

yep, the function is constantly 6 no matter how big x gets

Answer 24

Okay so what about something like

\[x->-\infty \frac{ 4x^2-3x-2-5x^3 }{ 9x^2+9x+7}\]

And it can't be factored, I tried

Answer 25

no need to factor as you are considering what happens as x approaches -infinity

Answer 26

what is the leading term in the numerator?

Answer 27

4

Answer 28

nope

Answer 29

it's not sorted in descending order

Answer 30

be careful to look out for this trick they often pull

Answer 31

Oh the answer would be either +- infinity

Answer 32

Because the power is higher in the numerator than the denominator

Answer 33

But I don't know how to find out which one it would be

Answer 34

I don't have to divide each term by x^3 do I? -___-

Answer 35

that's what I'd do

Answer 36

the term with the largest exponent is x^3

so divide everything by x^3

Answer 37

hmm hold on, let me think for a second

Answer 38

ok yeah, that trick works

Answer 39

hmm this site is being weird on me, idk if it's doing the same for you?

Answer 40

That's what my math teacher has us do ugh one moment let me do all that grunt work

Answer 41

alright

Answer 42

And since it's approaching -infinity and if we plugged that into x everything with x would just be 0 so my limit would be -5 right?

Answer 43

Or idk lol

Answer 44

it's not -5

Answer 45

what happened when you divided everything by x^3?

Answer 46

I got some stuff and fractions lol

I think I have to do what I didn't want to do and get common denominators etc?

Answer 47

Since I got

4/x-3/x^2-2/x^3-5/1 all divided by 9/x+9/x^2+7/x^3

Answer 48

So they're like double fractions. do I need to common denominate and then get like one term by doing reciprocal stuff?

Answer 49

this portion, 4/x-3/x^2-2/x^3, all goes to 0

why? because 4/x goes to 0 when x gets really large (in either direction +infinity or -infinity), 3/x^2 goes to 0, and 2/x^3 goes to zero. They all go to 0. Then you add up those zeros to get zero

Answer 50

same with 9/x+9/x^2+7/x^3

Answer 51

so we really have -5/0 when we simplify everything

Answer 52

which is the same as saying

limit as x --> infinity of -5/x

Answer 53

So x is approaching +infinity?

Answer 54

well in your problem

\[x->-\infty \frac{ 4x^2-3x-2-5x^3 }{ 9x^2+9x+7}\]

x is approaching -infinity

Answer 55

I can see it does from a graph, but I just don't know how to get it analytically..

So it would be -5/x which would be -5/-infinity which means the limit is +infinity

Answer 56

Since the negatives cancel?

Answer 57

yeah you can think of it like that

Answer 58

alright.. let's do another one of those lol

Answer 59

actually no wait

Answer 60

it's -5/0 not -5/infinity or -5/(-infinity)

Answer 61

So then how would I ever know what the limit is, it would be -5/0 which the limit DNE but he wants us to say something about it being +- infiinity. I can see when I graph it that its +infinity is the limit, but idk how to get that analytically.

Answer 62

Ohhhhhhhhhhhh my notes tell me to plug in very small or large numbers to see if it's + or 0 infinity, I got it now.

Answer 63

yeah that's using the table method more or less

Answer 64

Which equation would I plug it into though? My original or the one after I divided everything by x^3?

Answer 65

x is approaching negative infinity

in the denominator, we have 9/x+9/x^2+7/x^3. That denominator approaches 0 from the left. Why left? Because this portion is effectively approaching 0 AND it is approaching from the negative side due to the negative numbers in the denominator

----------------

So we this reduced expression -5/x

x is now approaching 0 from the left (reason stated above)

the limit for -5/x when x approaches 0 from the left is +infinity (plug in small values near zero to the left of 0)

Answer 66

we have*

Answer 67

and you would work with the reduced/simplified equation

Answer 68

or else, why reduce it if you aren't going to work with it?

Answer 69

That's true, I see,thanks! Let's do another though anyways just to make sure I got it lol

Answer 70

alright

Answer 71

\[Lim x->\infty \frac{ 3x^3-23 }{ 4x-1 }\]

Answer 72

ok go ahead and show what you have so far

Answer 73

So as x approaches infinity from the left we get - infinity and as it approaches it from the right we get +infinity?

Answer 74

So the limit as x approaches infinity DNE?

Answer 75

how do you approach infinity from the left? or right?

Answer 76

infinity is way out there that you can't touch it

Answer 77

Good question.

Answer 78

I just plugged in + or negative values idk what to do okay lol

Alls i know is I get 3/0 or 3/x

Answer 79

why not use that same trick we did last time

Answer 80

the leading term is 3x^3

so divide everything by x^3

Answer 81

So if I plug in a small number and a big numbers it's just always going to be positive.

Do I not plug in any negative values?

Answer 82

oh looks like you did that, nvm lol

Answer 83

so we have 3/x

Answer 84

let's backtrack a sec

the denominator turns from 4x-1 into 4x/x^3-1/x^3 after we divide everything by x^3

Answer 85

4x/x^3-1/x^3 approaches 0 as x ---> +infinity

key thing to remember: it approaches 0 from the right (the expression is getting smaller and smaller, but still remains positive)

Answer 86

that critical piece of info allows us to go back to 3/x

and we now have the limit as x ---> 0 from the right for 3/x

which gives +infinity

Answer 87

Yeah and to the left is -infinity

Answer 88

another way to look at this

the numerator has 3x^3 in it, which is the largest monomial. As x --> +infinity, x^3 is going to be the dominant term that tells you what is going on with the whole expression. If you cube a very large positive number, you get something even bigger and more positive

so that's why we approach +infinity

Answer 89

yes

Answer 90

Oh that's an easier way to look at it

Answer 91

Okay so piecewise functions...

Answer 92

So we have:

Find lim -> -1

x^3-1,x>=-1
2x,x<-1

Answer 93

Is it just -2...?

Answer 94

left hand limit: x^3 - 1 = (-1)^3 - 1 = -2

right hand limit: 2x = 2(-1) = -2

LHL = RHL = -2

so yep

Answer 95

because the limit exists at x = -1, and because f(-1) is also defined, this means the function is continuous at x = -1

Answer 96

Okay so, I don't remember how to factor x^3... lol.. So how would I do something like:

\[\lim x-> 1 \frac{ x^3+x^2-5x+3 }{ x^3-3x+2 }\]

Answer 97

oh sry, forgot to also mention that limit at -1 equals f(-1), which is the 3rd condition for continuity