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

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

Mathematics
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|dw:1433989546928:dw|
|dw:1433989546928:dw|\[(2x^2 -x -3)/ (2x^2 -5x +3)\]
limit is x -> 1 and theres a + beneath the limit. how do i do this?

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\[\lim_{x\rightarrow ^+1} \frac{ (2x^2−x−3)}{(2x^2−5x+3)}\] This?
So, are you familiar with what the + under the limit means?
@Anikate you still there? :)
@jtvatsim yea sorry had to use the bathroom
I do not know what that means. sorry fpr late respoce
You know that isn't allowed... jk :)
lolz
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.
is there such thing as a two sided?
Yes, most limits are two-sided. This is what is indicated by having no + or - sign under the limit.
oh ok, so what is the significance of the +?
The + restricts the limit to coming only from the right side of the graph, a picture will help.
|dw:1433990529989:dw|
hmm..... lost at that picture
In this picture we have, \[\lim_{x\rightarrow ^+3} graph = 2\] \[\lim_{x\rightarrow ^-3} graph = 1\] \[\lim_{x\rightarrow ^3} graph = DNE\]
As you approach x = 3 from the right the y values tend to 2.
As you approach x = 3 from the left the y values tend to 1.
If you approach x = 3 from both sides, you do not hit the same number. There is no two-sided limit in this picture.
what do u mean by grpah=2 or graph=1?
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... :) ).
oh ok
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.
and what about the "tend to 2"
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".
oh ok so whats next? do we grpah my equation to find the asnwer?
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).
oh ok, so how do do that?
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.
ok
\[\lim_{x\rightarrow ^+1} \frac{(2x^2−x−3)}{(2x^2−5x+3)} = \frac{2(1)^2-1-3}{2(1)^2-5+3}\]
i got 0
But where did you get the zero?
\[\frac{-2}{0}\]
yup... wait nvm, its UND
UNDEFINED, cuz the 0 is on the bottom
Right, this is either positive or negative infinity.
idk which one
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.
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.
all the x's we plugged in WERE 1
why 1.01? why not 2?
Well, yes I know, that's cuz we cheated what the limit actually tells us to do.
If we go too far we could potentially get false results.
And, yes, how do we define "too far" is the follow up to that.. :)
how come? doesnt the + mean anythign greater than 1?
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.
Welcome to the philosophy of Calculus. The precise science of imprecision and approximation where paradoxes happen left and right. :)
oh ok, so what if it was - instead of +, would we use .99?
Yep, that's the idea.
I mean it would be even better to use x = 1.0000000000001 in our current situation, but nobody wants to do that.
x = 1.01 should get us close enough.
ok, sounds good. whats next?
In any case, when we plug in x = 1.01, we get...
201
yup thats what i got
Assuming that this behavior is continuous, I think that we have good reason to believe that the true limit is positive infinity.
how did we tell it was POSTIVE and how Infinity?
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.
so any number divided by 0 is inifinity?
Any NONZERO number divided by 0 is positive or negative infinity. 0/0 is indeterminate and requires additional tests.
oh ok so the answer is positive infinity?
cn
Yes. This can also be confirmed via a graph. (Try Desmos). :)
can you please help me with one more? and i hav
a graphing clalc next to me haha
Yay!
|dw:1433992014016:dw|
Is that a superscript -, or an actual negative 3?
not even sure what im supposed to find an actual negative 3
Same question as before, find the limit as x goes to -3.
its an actual negative 3
is 0/0
So this one is very unfriendly... :)
haha ill try to u
rst
understand it
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.
havent heard of 2, but is it easier than 1?
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.
im in precalc, but if its easier pleace tell me :D
If you are in Precalc, it may not be easier yet, it depends on if you know what derivatives are?
sounds familiar, what does it look like real quick
is hospitals like a formula?
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?
nope
will it work if i dont know?
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
If you can memorize these facts, you will be able to use L'Hospital's rule.
got it, keep going
why 3x become 3(1), where did the 1 come form?
Because x always becomes 1 after applying the derivative.
o ok, gotcha
So, for instance, the equation x^3 + 27 becomes 3x^2 + 0 after applying the derivative.
gotcha
does the denominator become -4?
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".
Close, -x - 3 becomes -1 - 0 since any number becomes 0 after the deriviative.
oh yea
The derivative sort of disintegrates algebra equations into simpler and simpler equations.
By L'Hospital's Rule, we can say this:
-3x^2
\[\lim_{x\rightarrow -3} \frac{x^3 + 27}{-x-3} = \lim_{x\rightarrow -3} \frac{3x^2 + 0}{-1 - 0}=\lim_{x\rightarrow -3} -3x^2\]
Yes, you are right.
Now, if you plug in x = -3, you will get the answer.
I dont understand what my answers should look like?
Answers to limit questions are numbers or infinities.
why do some look like infinites and some jus regular numbers?
how do i decide if it will be a infinite or a number ?
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.
If you get a limit that says (number/0) the answer will be infinity (or Does Not Exist depending on your teacher).
If you get a limit that equals an actual number than the answer is that number.
what if i got -1 for one of them?
after plugging in the x value into the equation
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.
no after plugging in the x value and everythign my answer was -1, what do i do?
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.
oh ok gotcha, can you help me with a piecewise function?
Sure, and by the way our answer to this previous question is -27. Just checking.
yup i got -27
just a min i will post a pic
OK cool, your teachers will not expect you to use that L'Hospital method, but it just happened. :)
lol it will be fun surprising my math teacher with that lol math hacks
i didnt do the piece wise right, can u plz explain to me how to graph and get the answers?
OK, let's see...
So, would you know how to graph -x^2 + 1 if it was just by itself?
I just plug in random x values and get the y values to plot
OK, that works. The only thing you will do differently is only pick random x values that are less than 0.
That is what the piecewise function is telling you "Plot -x^2 + 1" for x values less than 0.
yea i didx that
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|
Alright, now we do the same thing with the |2x-1| for x >= 0 graph.
why would we plot at 0? do we really need it? i know the inequality says x<0 not lesser than or equal to
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.
oh ok, it cant just end on the x-axis at -1?
and leave a open circle there?
Only if you are considering whole numbers. I get the feeling that we should include all decimal numbers as well.
That's the standard practice anyways.
why decimals ?
Well, certainly x = -0.0000001 is less than 0 and should be included as part of the graph
yea
oh so, we just start at 0 to be friendly?
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.
got it, how do we do the next peice wise functions?
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.
yup quick question
the absolute values
does that automatically turn the equation into 2x+1
Very good question.
Try a few numbers and see what you think... :)
or does the answer just turn positive if its negative?
What you just said is the best way to start. Take numbers, plug in, and make them positive if they are negative.
so does the equation turn postive from the beginning itself, or do we wait for the answer then turn it positive?
Wait for the answer. The absolute value bars are like parentheses and don't act until after the inner answer.
oh ok
You've had very good questions so far! You have the making of a great mathematician!
:D thanks!!
ok so theres an issue here
when i plot x=0 and x=1 it appears to be the same y value? looks odd on the graph
That feels weird right? But the math doesn't lie. You have it correct so far. :)
so should i make it a curved line or a straight line through the points?
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. :)
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.
oh yea i plotted it, i see it now
|dw:1433994694244:dw| Perfect!
now what are these conintues or dis continues things?
and jump discont etc,
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".
Here are two examples.
|dw:1433994827549:dw|
A discontinuous graph is "broken" in some sense.
mhm
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.
oh ok
Does that make sense? How about does the limit exist?
what do u mean by does the lmit exist?
yea idk how to do limits or anything like those on the piecewise functions
OK, so here's a quick test to tell if a limit exists.
ok
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.
Now, trace the graph with your fingers (moving them towards each other) going to x = 0.
pencil graph?
The graph you drew.
oh ok
done
If they touch when you reach x = 0, then the limit exists at x = 0.
any other way of checking other than that
Yes, as long as there are no jump discontinuities at the point, the limit exists.
This is exactly what the tracing trick tests for.
oh ok, so how do i say that a limit exists?
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.
yea he does, but just in case, is there any other way of saying it?
can i just say yes a limit exists, cuz it only
asks if a limit exists
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.
lol jsut a sec trying to translate all that in my head
lol, I guess that is a bit much... :)
so x=1?
The limit is 1. (The limit talks about y-values, if that helps).
So as x approaches 0 (x = 0), y goes to 1, (y = 1).
how do we tell if the limit is talking the x value or the y value?
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.
oh o k
If you like symbols, you could write \[\lim_{x\rightarrow 0}f(x) = 1\]
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.
oh gotcha
wait wh
what wou
be the answer to part a?
ugh sorry if my sentences break apart in messages, my laptop has an oddly placed mousepad hahah
no worries, I had to find the graph back.
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...)
That should be \[\lim_{x\rightarrow 0^-}\] sorry
oooh ok gotcha, and b.) is yes c.) is continuous
why cant it be lim f(x)=1?
without any x information underneath you mean?
no with a x info, but why is it lim, why cant we add the f(x)=1?
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?
noo
you said to write lim(underneath "x -> 0")
why cant I write lim f(x)=1 (underneath "x -> 0")?
OH! No that's right, you are doing it fine! I was being lazy on that reply. Teaches me... :)
oh ok, lol just makin sure, cuz u corrected urself and I was wondering why haha
so c.) continuous? right?
Yeah, I correct that I had written lim(x->1) It should have been going to 0.
Yes, your answers to b and c are correct.
ok, now how do I do D?
Well, is it discontinuous?
nope
So, there you have it. Just say, "Not discontinuous."
what are all the type of discontinuities?
You have "hole" (a missing value), "jump" (a gap), and "infinite" (an asymptote) where the graph shoots up or down to infinity.
thats the only 3 right? are there any different types of continuities?
There are different forms of continuity, but that is usually not discussed until advanced calculus at university.
oh thank goodness lol
lol, the first good news all night! :)
can we go over one more?
wait actually 2? second one is easy hopefully
do u live east coast USA?
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. :)
Nope, west coast.
OH! what timei s it for you?
9:15 or so.
should I open a new question, are u gonna join it?
Yeah, I'll join it. :)
just @ me.

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