need help with limits, posting an equation. all help is appreciated. medals...

- Anikate

need help with limits, posting an equation. all help is appreciated. medals...

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- Anikate

|dw:1433989546928:dw|

- Anikate

|dw:1433989546928:dw|\[(2x^2 -x -3)/ (2x^2 -5x +3)\]

- Anikate

limit is x -> 1 and theres a + beneath the limit. how do i do this?

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## More answers

- jtvatsim

\[\lim_{x\rightarrow ^+1} \frac{ (2x^2−x−3)}{(2x^2−5x+3)}\]
This?

- jtvatsim

So, are you familiar with what the + under the limit means?

- jtvatsim

@Anikate you still there? :)

- Anikate

@jtvatsim yea sorry had to use the bathroom

- Anikate

I do not know what that means. sorry fpr late respoce

- jtvatsim

You know that isn't allowed... jk :)

- Anikate

lolz

- jtvatsim

OK, so the +1 just means that the limit is approaching 1 from the right side of the graph. It is a "one-sided limit" if you've heard of that term.

- Anikate

is there such thing as a two sided?

- Anikate

@jtvatsim

- jtvatsim

Yes, most limits are two-sided. This is what is indicated by having no + or - sign under the limit.

- Anikate

oh ok, so what is the significance of the +?

- jtvatsim

The + restricts the limit to coming only from the right side of the graph, a picture will help.

- jtvatsim

|dw:1433990529989:dw|

- Anikate

hmm..... lost at that picture

- jtvatsim

In this picture we have,
\[\lim_{x\rightarrow ^+3} graph = 2\]
\[\lim_{x\rightarrow ^-3} graph = 1\]
\[\lim_{x\rightarrow ^3} graph = DNE\]

- jtvatsim

As you approach x = 3 from the right the y values tend to 2.

- jtvatsim

As you approach x = 3 from the left the y values tend to 1.

- jtvatsim

If you approach x = 3 from both sides, you do not hit the same number. There is no two-sided limit in this picture.

- Anikate

what do u mean by grpah=2 or graph=1?

- jtvatsim

I mean whatever the equation of that graph is. I'm using the word "graph" to stand for whatever the algebraic equation should be (I have no idea... :) ).

- Anikate

oh ok

- jtvatsim

In your case, you have been given the equation of some graph. And they are asking you to find out what the limit is from the right side.

- Anikate

and what about the "tend to 2"

- jtvatsim

That's how I understand the concept of a limit. The y values never actually become 2, they just approach it "arbitrarily close." Those words are weird to me, so I think about it as the y values "approach 2" or "tend to become 2".

- Anikate

oh ok so whats next?
do we grpah my equation to find the asnwer?

- jtvatsim

That is a very valid approach. We don't have to do that. If you recall, many limits can be found by simply plugging in the number that x is approaching. We might want to try that first (it's easier).

- Anikate

oh ok, so how do do that?

- jtvatsim

Well, we are looking for
\[\lim_{x\rightarrow ^+1} \frac{(2x^2−x−3)}{(2x^2−5x+3)}\]
Let's try substituting x = 1 into the equation.

- Anikate

ok

- jtvatsim

\[\lim_{x\rightarrow ^+1} \frac{(2x^2−x−3)}{(2x^2−5x+3)} = \frac{2(1)^2-1-3}{2(1)^2-5+3}\]

- Anikate

i got 0

- jtvatsim

But where did you get the zero?

- jtvatsim

\[\frac{-2}{0}\]

- Anikate

yup... wait nvm, its UND

- Anikate

UNDEFINED, cuz the 0 is on the bottom

- jtvatsim

Right, this is either positive or negative infinity.

- Anikate

idk which one

- jtvatsim

Now is where that +1 thing comes in handy. We know that all the x's we plugged in are just a little bit bigger than 1. They are to the right of 1 after all.

- jtvatsim

So, it might be best to test a number just a little bigger than 1 to see what kind of answer we get. Let's try x = 1.01.

- Anikate

all the x's we plugged in WERE 1

- Anikate

why 1.01? why not 2?

- jtvatsim

Well, yes I know, that's cuz we cheated what the limit actually tells us to do.

- jtvatsim

If we go too far we could potentially get false results.

- jtvatsim

And, yes, how do we define "too far" is the follow up to that.. :)

- Anikate

how come? doesnt the + mean anythign greater than 1?

- jtvatsim

Yes, but the limit is asking us to approach 1 as close as possible. x = 2 will approximate the limit, but it will probably be a bad approximation.

- jtvatsim

Welcome to the philosophy of Calculus. The precise science of imprecision and approximation where paradoxes happen left and right. :)

- Anikate

oh ok, so what if it was - instead of +, would we use .99?

- jtvatsim

Yep, that's the idea.

- jtvatsim

I mean it would be even better to use x = 1.0000000000001 in our current situation, but nobody wants to do that.

- jtvatsim

x = 1.01 should get us close enough.

- Anikate

ok, sounds good. whats next?

- jtvatsim

In any case, when we plug in x = 1.01, we get...

- jtvatsim

201

- Anikate

yup thats what i got

- jtvatsim

Assuming that this behavior is continuous, I think that we have good reason to believe that the true limit is positive infinity.

- Anikate

how did we tell it was POSTIVE and how Infinity?

- jtvatsim

I determined that it had to be either positive or negative infinity based on our first result (a number divided by 0). Then, since we got a positive answer just to the right of 1 (+201) it is reasonable to believe that the true answer is positive infinity.

- Anikate

so any number divided by 0 is inifinity?

- jtvatsim

Any NONZERO number divided by 0 is positive or negative infinity. 0/0 is indeterminate and requires additional tests.

- Anikate

oh ok so the answer is positive infinity?

- Anikate

cn

- jtvatsim

Yes. This can also be confirmed via a graph. (Try Desmos). :)

- Anikate

can you please help me with one more? and i hav

- Anikate

a graphing clalc next to me haha

- jtvatsim

Yay!

- Anikate

|dw:1433992014016:dw|

- jtvatsim

Is that a superscript -, or an actual negative 3?

- Anikate

not even sure what im supposed to find an actual negative 3

- jtvatsim

Same question as before, find the limit as x goes to -3.

- Anikate

its an actual negative 3

- Anikate

is 0/0

- jtvatsim

So this one is very unfriendly... :)

- Anikate

haha ill try to u

- Anikate

rst

- Anikate

understand it

- jtvatsim

In these situations we have two options: 1) factor or 2) use L'Hospital's Rule. If you haven't heard of 2 yet, then ignore that.

- Anikate

havent heard of 2, but is it easier than 1?

- jtvatsim

Yes, but it usually isn't introduced until Calc II. Most Calc I teachers don't like to give it away early, but we can do it both ways if you want.

- Anikate

im in precalc, but if its easier pleace tell me :D

- jtvatsim

If you are in Precalc, it may not be easier yet, it depends on if you know what derivatives are?

- Anikate

sounds familiar, what does it look like real quick

- Anikate

is hospitals like a formula?

- jtvatsim

Yes, but it requires you to know derivatives well. For example, if you have x^2, then the derivative is 2x. Is that something you've done?

- Anikate

nope

- Anikate

will it work if i dont know?

- jtvatsim

Well, I would have to tell you the specific derivatives for this equation, but you would probably not be able to use it on different questions:
In short, derivatives change one algebra equation into another, they work like this (a short intro):
Any number -> becomes 0
x -> becomes 1
x^2 -> becomes 2x
x^3 -> becomes 3x^2
x^4 -> becomes 4x^3
Multiplies of expressions -> become multiples of derivatives
example
3x -> becomes 3(1) = 3
4x^2 -> becomes 4(2x) = 8x

- jtvatsim

If you can memorize these facts, you will be able to use L'Hospital's rule.

- Anikate

got it, keep going

- Anikate

why 3x become 3(1), where did the 1 come form?

- jtvatsim

Because x always becomes 1 after applying the derivative.

- Anikate

o ok, gotcha

- jtvatsim

So, for instance, the equation x^3 + 27 becomes 3x^2 + 0 after applying the derivative.

- Anikate

gotcha

- Anikate

does the denominator become -4?

- jtvatsim

How does this help? Well L'Hospital's Rule says this:
"If you take a limit and the answer is 0/0, you may apply the derivative to the top and bottom equation and the answer will be equivalent".

- jtvatsim

Close, -x - 3 becomes -1 - 0 since any number becomes 0 after the deriviative.

- Anikate

oh yea

- jtvatsim

The derivative sort of disintegrates algebra equations into simpler and simpler equations.

- jtvatsim

By L'Hospital's Rule, we can say this:

- Anikate

-3x^2

- jtvatsim

\[\lim_{x\rightarrow -3} \frac{x^3 + 27}{-x-3} = \lim_{x\rightarrow -3} \frac{3x^2 + 0}{-1 - 0}=\lim_{x\rightarrow -3} -3x^2\]

- jtvatsim

Yes, you are right.

- jtvatsim

Now, if you plug in x = -3, you will get the answer.

- Anikate

I dont understand what my answers should look like?

- jtvatsim

Answers to limit questions are numbers or infinities.

- Anikate

why do some look like infinites and some jus regular numbers?

- Anikate

how do i decide if it will be a infinite or a number
?

- jtvatsim

Well, those are all the possible numbers that exist. Sometimes y values shoot up very fast into infinity land (like the previous graph). Other times y values level off and reach an actual number.

- jtvatsim

If you get a limit that says (number/0) the answer will be infinity (or Does Not Exist depending on your teacher).

- jtvatsim

If you get a limit that equals an actual number than the answer is that number.

- Anikate

what if i got -1 for one of them?

- Anikate

after plugging in the x value into the equation

- jtvatsim

Then just treat it like dividing by -1, I think you are talking about what we just did above. Nothing special happens. We know what to do when we see a -1 dividing. We have no idea what to do if we see a 0 dividing.

- Anikate

no after plugging in the x value and everythign my answer was -1, what do i do?

- jtvatsim

If after plugging in the x value your answer is -1, then that is the answer to the limit question. You would write:
lim_x-> whatever (some equation) = -1.

- Anikate

oh ok gotcha, can you help me with a piecewise function?

- jtvatsim

Sure, and by the way our answer to this previous question is -27. Just checking.

- Anikate

yup i got -27

- Anikate

just a min i will post a pic

- jtvatsim

OK cool, your teachers will not expect you to use that L'Hospital method, but it just happened. :)

- Anikate

##### 1 Attachment

- Anikate

lol it will be fun surprising my math teacher with that lol math hacks

- Anikate

i didnt do the piece wise right, can u plz explain to me how to graph and get the answers?

- jtvatsim

OK, let's see...

- jtvatsim

So, would you know how to graph -x^2 + 1 if it was just by itself?

- Anikate

I just plug in random x values and get the y values to plot

- jtvatsim

OK, that works. The only thing you will do differently is only pick random x values that are less than 0.

- jtvatsim

That is what the piecewise function is telling you "Plot -x^2 + 1" for x values less than 0.

- Anikate

yea i didx that

- jtvatsim

Good. We can imagine that at x = 0 (right on the not allowed zone) we would get a value of 1, so something like this picture ...|dw:1433993796086:dw|

- jtvatsim

Alright, now we do the same thing with the |2x-1| for x >= 0 graph.

- Anikate

why would we plot at 0? do we really need it? i know the inequality says x<0 not lesser than or equal to

- jtvatsim

We don't "really" need it, it's just a nice visual barrier to have an 'open dot' at 0. As you say, there isn't really anything plotted there, it is just a reminder of where the plot ends.

- Anikate

oh ok, it cant just end on the x-axis at -1?

- Anikate

and leave a open circle there?

- jtvatsim

Only if you are considering whole numbers. I get the feeling that we should include all decimal numbers as well.

- jtvatsim

That's the standard practice anyways.

- Anikate

why decimals
?

- jtvatsim

Well, certainly x = -0.0000001 is less than 0 and should be included as part of the graph

- Anikate

yea

- Anikate

oh so, we just start at 0 to be friendly?

- jtvatsim

Right. This gives us a feeling for the shape of the graph. Since, of course, you can't plug in all the decimals that are out there, so we end up connecting the dots just to get a sense of the shape.

- Anikate

got it, how do we do the next peice wise functions?

- jtvatsim

Well, let's use the same strategy since it worked for the last one. Now we are allowed to plug in x = 0 and greater.

- Anikate

yup quick question

- Anikate

the absolute values

- Anikate

does that automatically turn the equation into 2x+1

- jtvatsim

Very good question.

- jtvatsim

Try a few numbers and see what you think... :)

- Anikate

or does the answer just turn positive if its negative?

- jtvatsim

What you just said is the best way to start. Take numbers, plug in, and make them positive if they are negative.

- Anikate

so does the equation turn postive from the beginning itself, or do we wait for the answer then turn it positive?

- jtvatsim

Wait for the answer. The absolute value bars are like parentheses and don't act until after the inner answer.

- Anikate

oh ok

- jtvatsim

You've had very good questions so far! You have the making of a great mathematician!

- Anikate

:D thanks!!

- Anikate

ok so theres an issue here

- Anikate

when i plot x=0 and x=1 it appears to be the same y value? looks odd on the graph

- jtvatsim

That feels weird right? But the math doesn't lie. You have it correct so far. :)

- Anikate

so should i make it a curved line or a straight line through the points?

- jtvatsim

Good question. You could confirm by graphing points like x = 1/2, x = 1/4, and x = 3/4, but I won't torture you. :)

- jtvatsim

The inner equation is that of a line, 2x - 1. The absolute value graph will act the same except it will be the shape of a "V" now. Here's a picture.

- Anikate

oh yea i plotted it, i see it now

- jtvatsim

|dw:1433994694244:dw|
Perfect!

- Anikate

now what are these conintues or dis continues things?

- Anikate

and jump discont etc,

- jtvatsim

Continuous functions are basically those that you could draw without lifting your pencil. In other words, they have no "holes" and no "gaps" or "jumps".

- jtvatsim

Here are two examples.

- jtvatsim

|dw:1433994827549:dw|

- jtvatsim

A discontinuous graph is "broken" in some sense.

- Anikate

mhm

- jtvatsim

When we first plotted -x^2 + 1, we had a hole at x = 0, y = 1 since we weren't really allowed to plot there. However, after we plotted |2x - 1| we filled that exact hole in with x = 0, y = 1 which was allowed to plot. There is no hole at x = 0, obviously no jump, so the graph is continuous at x = 0.

- Anikate

oh ok

- jtvatsim

Does that make sense? How about does the limit exist?

- Anikate

what do u mean by does the lmit exist?

- Anikate

yea idk how to do limits or anything like those on the piecewise functions

- jtvatsim

OK, so here's a quick test to tell if a limit exists.

- Anikate

ok

- jtvatsim

Take your right finger and place it on the right side of the pencil graph. Take your left finger and place it on the left side of the pencil graph.

- jtvatsim

Now, trace the graph with your fingers (moving them towards each other) going to x = 0.

- Anikate

pencil graph?

- jtvatsim

The graph you drew.

- Anikate

oh ok

- Anikate

done

- jtvatsim

If they touch when you reach x = 0, then the limit exists at x = 0.

- Anikate

any other way of checking other than that

- jtvatsim

Yes, as long as there are no jump discontinuities at the point, the limit exists.

- jtvatsim

This is exactly what the tracing trick tests for.

- Anikate

oh ok, so how do i say that a limit exists?

- jtvatsim

Your teacher probably will say something like:
"The limit exists when the left-hand limit equals the right-hand limit"
Again, this is the same thing as the tracing trick.

- Anikate

yea he does, but just in case, is there any other way of saying it?

- Anikate

can i just say yes a limit exists, cuz it only

- Anikate

asks if a limit exists

- jtvatsim

Sure. It's a good practice to give your reason for saying that. You could simply say, the left-hand limit as x approaches 0 is 1, and the right-hand limit as x approaches 0 is 1. Since they are equal, the limit as x approaches 0 exists and is equal to 1.

- Anikate

lol jsut a sec trying to translate all that in my head

- jtvatsim

lol, I guess that is a bit much... :)

- Anikate

so x=1?

- jtvatsim

The limit is 1. (The limit talks about y-values, if that helps).

- jtvatsim

So as x approaches 0 (x = 0), y goes to 1, (y = 1).

- Anikate

how do we tell if the limit is talking the x value or the y value?

- jtvatsim

In general, the limit always talks about the y value. I suppose there are cases where it could talk about the x value, but this is not standard practice.

- Anikate

oh o k

- jtvatsim

If you like symbols, you could write \[\lim_{x\rightarrow 0}f(x) = 1\]

- jtvatsim

f(x) is the piecewise function (that produces y values). Our x is moving toward 0, so we expect that y will end up being 1.

- Anikate

oh gotcha

- Anikate

wait wh

- Anikate

what wou

- Anikate

be the answer to part a?

- Anikate

ugh sorry if my sentences break apart in messages, my laptop has an oddly placed mousepad hahah

- jtvatsim

no worries, I had to find the graph back.

- jtvatsim

OK, so the left-hand limit would be just tracing the graph with your left finger until you get to x = 0 and report the y value you get (it will be 1). You would write:
\[\lim_{x\rightarrow 1^-} f(x) = 1\] Note the - sign to indicate from the left (if you ask me L would make more sense, but...)

- jtvatsim

That should be \[\lim_{x\rightarrow 0^-}\] sorry

- Anikate

oooh ok gotcha, and b.) is yes c.) is continuous

- Anikate

why cant it be lim f(x)=1?

- jtvatsim

without any x information underneath you mean?

- Anikate

no with a x info, but why is it lim, why cant we add the f(x)=1?

- jtvatsim

So, you are saying why make this complicated by saying lim f(x) = 1? Why not just say f(x) = 1 when x = 0 instead?

- Anikate

noo

- Anikate

you said to write lim(underneath "x -> 0")

- Anikate

why cant I write lim f(x)=1 (underneath "x -> 0")?

- jtvatsim

OH! No that's right, you are doing it fine! I was being lazy on that reply. Teaches me... :)

- Anikate

oh ok, lol just makin sure, cuz u corrected urself and I was wondering why haha

- Anikate

so c.) continuous? right?

- jtvatsim

Yeah, I correct that I had written lim(x->1) It should have been going to 0.

- jtvatsim

Yes, your answers to b and c are correct.

- Anikate

ok, now how do I do D?

- jtvatsim

Well, is it discontinuous?

- Anikate

nope

- jtvatsim

So, there you have it. Just say, "Not discontinuous."

- Anikate

what are all the type of discontinuities?

- jtvatsim

You have "hole" (a missing value), "jump" (a gap), and "infinite" (an asymptote) where the graph shoots up or down to infinity.

- Anikate

thats the only 3 right? are there any different types of continuities?

- jtvatsim

There are different forms of continuity, but that is usually not discussed until advanced calculus at university.

- Anikate

oh thank goodness lol

- jtvatsim

lol, the first good news all night! :)

- Anikate

can we go over one more?

- Anikate

wait actually 2? second one is easy hopefully

- Anikate

do u live east coast USA?

- jtvatsim

Go ahead and close this question. I may have to run, but if you post a new question, others can see it. I'll stick around as long as possible though. :)

- jtvatsim

Nope, west coast.

- Anikate

OH! what timei s it for you?

- jtvatsim

9:15 or so.

- Anikate

should I open a new question, are u gonna join it?

- jtvatsim

Yeah, I'll join it. :)

- jtvatsim

just @ me.

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