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Anikate
 one year ago
need help with limits, posting an equation. all help is appreciated. medals...
Anikate
 one year ago
need help with limits, posting an equation. all help is appreciated. medals...

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Anikate
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433989546928:dw\[(2x^2 x 3)/ (2x^2 5x +3)\]

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0limit is x > 1 and theres a + beneath the limit. how do i do this?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6\[\lim_{x\rightarrow ^+1} \frac{ (2x^2−x−3)}{(2x^2−5x+3)}\] This?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6So, are you familiar with what the + under the limit means?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6@Anikate you still there? :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0@jtvatsim yea sorry had to use the bathroom

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0I do not know what that means. sorry fpr late respoce

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6You know that isn't allowed... jk :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6OK, so the +1 just means that the limit is approaching 1 from the right side of the graph. It is a "onesided limit" if you've heard of that term.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0is there such thing as a two sided?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yes, most limits are twosided. This is what is indicated by having no + or  sign under the limit.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, so what is the significance of the +?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6The + restricts the limit to coming only from the right side of the graph, a picture will help.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6dw:1433990529989:dw

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0hmm..... lost at that picture

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6In this picture we have, \[\lim_{x\rightarrow ^+3} graph = 2\] \[\lim_{x\rightarrow ^3} graph = 1\] \[\lim_{x\rightarrow ^3} graph = DNE\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6As you approach x = 3 from the right the y values tend to 2.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6As you approach x = 3 from the left the y values tend to 1.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6If you approach x = 3 from both sides, you do not hit the same number. There is no twosided limit in this picture.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0what do u mean by grpah=2 or graph=1?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6I 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... :) ).

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6In 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
 one year ago
Best ResponseYou've already chosen the best response.0and what about the "tend to 2"

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6That'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
 one year ago
Best ResponseYou've already chosen the best response.0oh ok so whats next? do we grpah my equation to find the asnwer?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6That 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
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, so how do do that?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Well, 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.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6\[\lim_{x\rightarrow ^+1} \frac{(2x^2−x−3)}{(2x^2−5x+3)} = \frac{2(1)^213}{2(1)^25+3}\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6But where did you get the zero?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0yup... wait nvm, its UND

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0UNDEFINED, cuz the 0 is on the bottom

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Right, this is either positive or negative infinity.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Now 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
 one year ago
Best ResponseYou've already chosen the best response.6So, 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
 one year ago
Best ResponseYou've already chosen the best response.0all the x's we plugged in WERE 1

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Well, yes I know, that's cuz we cheated what the limit actually tells us to do.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6If we go too far we could potentially get false results.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6And, yes, how do we define "too far" is the follow up to that.. :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0how come? doesnt the + mean anythign greater than 1?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.6Welcome to the philosophy of Calculus. The precise science of imprecision and approximation where paradoxes happen left and right. :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, so what if it was  instead of +, would we use .99?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yep, that's the idea.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6I mean it would be even better to use x = 1.0000000000001 in our current situation, but nobody wants to do that.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6x = 1.01 should get us close enough.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0ok, sounds good. whats next?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6In any case, when we plug in x = 1.01, we get...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Assuming that this behavior is continuous, I think that we have good reason to believe that the true limit is positive infinity.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0how did we tell it was POSTIVE and how Infinity?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6I 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
 one year ago
Best ResponseYou've already chosen the best response.0so any number divided by 0 is inifinity?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Any NONZERO number divided by 0 is positive or negative infinity. 0/0 is indeterminate and requires additional tests.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0oh ok so the answer is positive infinity?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yes. This can also be confirmed via a graph. (Try Desmos). :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0can you please help me with one more? and i hav

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0a graphing clalc next to me haha

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Is that a superscript , or an actual negative 3?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0not even sure what im supposed to find an actual negative 3

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Same question as before, find the limit as x goes to 3.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0its an actual negative 3

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6So this one is very unfriendly... :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6In 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
 one year ago
Best ResponseYou've already chosen the best response.0havent heard of 2, but is it easier than 1?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0im in precalc, but if its easier pleace tell me :D

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6If you are in Precalc, it may not be easier yet, it depends on if you know what derivatives are?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0sounds familiar, what does it look like real quick

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0is hospitals like a formula?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0will it work if i dont know?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Well, 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
 one year ago
Best ResponseYou've already chosen the best response.6If you can memorize these facts, you will be able to use L'Hospital's rule.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0why 3x become 3(1), where did the 1 come form?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Because x always becomes 1 after applying the derivative.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6So, for instance, the equation x^3 + 27 becomes 3x^2 + 0 after applying the derivative.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0does the denominator become 4?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6How 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
 one year ago
Best ResponseYou've already chosen the best response.6Close, x  3 becomes 1  0 since any number becomes 0 after the deriviative.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6The derivative sort of disintegrates algebra equations into simpler and simpler equations.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6By L'Hospital's Rule, we can say this:

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6\[\lim_{x\rightarrow 3} \frac{x^3 + 27}{x3} = \lim_{x\rightarrow 3} \frac{3x^2 + 0}{1  0}=\lim_{x\rightarrow 3} 3x^2\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Now, if you plug in x = 3, you will get the answer.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0I dont understand what my answers should look like?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Answers to limit questions are numbers or infinities.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0why do some look like infinites and some jus regular numbers?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0how do i decide if it will be a infinite or a number ?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Well, 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
 one year ago
Best ResponseYou've already chosen the best response.6If you get a limit that says (number/0) the answer will be infinity (or Does Not Exist depending on your teacher).

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6If you get a limit that equals an actual number than the answer is that number.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0what if i got 1 for one of them?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0after plugging in the x value into the equation

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Then 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
 one year ago
Best ResponseYou've already chosen the best response.0no after plugging in the x value and everythign my answer was 1, what do i do?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6If 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
 one year ago
Best ResponseYou've already chosen the best response.0oh ok gotcha, can you help me with a piecewise function?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Sure, and by the way our answer to this previous question is 27. Just checking.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0just a min i will post a pic

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6OK cool, your teachers will not expect you to use that L'Hospital method, but it just happened. :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0lol it will be fun surprising my math teacher with that lol math hacks

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0i didnt do the piece wise right, can u plz explain to me how to graph and get the answers?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6So, would you know how to graph x^2 + 1 if it was just by itself?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0I just plug in random x values and get the y values to plot

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6OK, that works. The only thing you will do differently is only pick random x values that are less than 0.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6That is what the piecewise function is telling you "Plot x^2 + 1" for x values less than 0.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Good. 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
 one year ago
Best ResponseYou've already chosen the best response.6Alright, now we do the same thing with the 2x1 for x >= 0 graph.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0why would we plot at 0? do we really need it? i know the inequality says x<0 not lesser than or equal to

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6We 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
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, it cant just end on the xaxis at 1?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0and leave a open circle there?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Only if you are considering whole numbers. I get the feeling that we should include all decimal numbers as well.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6That's the standard practice anyways.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Well, certainly x = 0.0000001 is less than 0 and should be included as part of the graph

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0oh so, we just start at 0 to be friendly?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Right. 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
 one year ago
Best ResponseYou've already chosen the best response.0got it, how do we do the next peice wise functions?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Well, 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
 one year ago
Best ResponseYou've already chosen the best response.0does that automatically turn the equation into 2x+1

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Try a few numbers and see what you think... :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0or does the answer just turn positive if its negative?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6What you just said is the best way to start. Take numbers, plug in, and make them positive if they are negative.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0so does the equation turn postive from the beginning itself, or do we wait for the answer then turn it positive?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Wait for the answer. The absolute value bars are like parentheses and don't act until after the inner answer.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6You've had very good questions so far! You have the making of a great mathematician!

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0ok so theres an issue here

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0when i plot x=0 and x=1 it appears to be the same y value? looks odd on the graph

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6That feels weird right? But the math doesn't lie. You have it correct so far. :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0so should i make it a curved line or a straight line through the points?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Good 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
 one year ago
Best ResponseYou've already chosen the best response.6The 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
 one year ago
Best ResponseYou've already chosen the best response.0oh yea i plotted it, i see it now

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6dw:1433994694244:dw Perfect!

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0now what are these conintues or dis continues things?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Continuous 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
 one year ago
Best ResponseYou've already chosen the best response.6Here are two examples.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6dw:1433994827549:dw

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6A discontinuous graph is "broken" in some sense.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6When 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.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Does that make sense? How about does the limit exist?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0what do u mean by does the lmit exist?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0yea idk how to do limits or anything like those on the piecewise functions

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6OK, so here's a quick test to tell if a limit exists.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Take 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
 one year ago
Best ResponseYou've already chosen the best response.6Now, trace the graph with your fingers (moving them towards each other) going to x = 0.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6If they touch when you reach x = 0, then the limit exists at x = 0.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0any other way of checking other than that

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yes, as long as there are no jump discontinuities at the point, the limit exists.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6This is exactly what the tracing trick tests for.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, so how do i say that a limit exists?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Your teacher probably will say something like: "The limit exists when the lefthand limit equals the righthand limit" Again, this is the same thing as the tracing trick.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0yea he does, but just in case, is there any other way of saying it?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0can i just say yes a limit exists, cuz it only

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0asks if a limit exists

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Sure. It's a good practice to give your reason for saying that. You could simply say, the lefthand limit as x approaches 0 is 1, and the righthand limit as x approaches 0 is 1. Since they are equal, the limit as x approaches 0 exists and is equal to 1.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0lol jsut a sec trying to translate all that in my head

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6lol, I guess that is a bit much... :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6The limit is 1. (The limit talks about yvalues, if that helps).

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6So as x approaches 0 (x = 0), y goes to 1, (y = 1).

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0how do we tell if the limit is talking the x value or the y value?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6In 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.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6If you like symbols, you could write \[\lim_{x\rightarrow 0}f(x) = 1\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6f(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
 one year ago
Best ResponseYou've already chosen the best response.0be the answer to part a?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0ugh sorry if my sentences break apart in messages, my laptop has an oddly placed mousepad hahah

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6no worries, I had to find the graph back.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6OK, so the lefthand 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
 one year ago
Best ResponseYou've already chosen the best response.6That should be \[\lim_{x\rightarrow 0^}\] sorry

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0oooh ok gotcha, and b.) is yes c.) is continuous

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0why cant it be lim f(x)=1?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6without any x information underneath you mean?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0no with a x info, but why is it lim, why cant we add the f(x)=1?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6So, 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
 one year ago
Best ResponseYou've already chosen the best response.0you said to write lim(underneath "x > 0")

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0why cant I write lim f(x)=1 (underneath "x > 0")?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6OH! No that's right, you are doing it fine! I was being lazy on that reply. Teaches me... :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, lol just makin sure, cuz u corrected urself and I was wondering why haha

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0so c.) continuous? right?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yeah, I correct that I had written lim(x>1) It should have been going to 0.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yes, your answers to b and c are correct.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0ok, now how do I do D?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Well, is it discontinuous?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6So, there you have it. Just say, "Not discontinuous."

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0what are all the type of discontinuities?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6You have "hole" (a missing value), "jump" (a gap), and "infinite" (an asymptote) where the graph shoots up or down to infinity.

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0thats the only 3 right? are there any different types of continuities?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6There are different forms of continuity, but that is usually not discussed until advanced calculus at university.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6lol, the first good news all night! :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0can we go over one more?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0wait actually 2? second one is easy hopefully

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0do u live east coast USA?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Go 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. :)

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0OH! what timei s it for you?

Anikate
 one year ago
Best ResponseYou've already chosen the best response.0should I open a new question, are u gonna join it?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.6Yeah, I'll join it. :)
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