Anikate
  • Anikate
need help with limits, posting an equation. all help is appreciated. medals...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Anikate
  • Anikate
|dw:1433989546928:dw|
Anikate
  • Anikate
|dw:1433989546928:dw|\[(2x^2 -x -3)/ (2x^2 -5x +3)\]
Anikate
  • Anikate
limit is x -> 1 and theres a + beneath the limit. how do i do this?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.