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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?

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

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

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

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

lolz

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

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.

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

oh ok

and what about the "tend to 2"

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

oh ok, so how do do that?

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

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?

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

Yep, that's the idea.

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

how did we tell it was POSTIVE and how Infinity?

so any number divided by 0 is inifinity?

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

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

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?

nope

will it work if i dont know?

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?

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

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
?

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

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

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

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

yea i didx that

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

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

and leave a open circle there?

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?

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

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?

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?

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,

Here are two examples.

|dw:1433994827549:dw|

A discontinuous graph is "broken" in some sense.

mhm

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

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?

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

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?

oh o k

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

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.

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?

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?

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

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?

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.