I am mr. Tarrou. In this calculus lesson

we’re gonna be doing two examples of finding limits of functions that involve

absolute value functions within them this is going to be a continuation of

course of us finding limits using properties of limits if you want more

examples there’ll be a link in the description of this video

but I saw these problems recently and thought they were rather interesting and

not something I’ve done in the past so one thing you’re going to see over and

over again through this year that as we deal with functions that involve

absolute values we’re going to be setting up piecewise functions because

of course you know an absolute value function tells you how far a number is

away from zero on the number line and if you have a positive number like the

absolute value of 5 well you count back your 5 units away

from the number from 0 on the number line and the absolute value negative 5

well the absolute value negative 5 is 5 units away from 0 on the number line

even though I think I just move my hand 4 times but you know you get the point

so we’re going to be through limits through derivatives and through

integration we’re going to be rewriting these problems in terms of a piecewise

function you can also say that it’s also the square root of something squared I

have a of a derivative video like that and later as I’ve had more experience

with calculus teaching it to my students the piecewise idea is much simpler so

we’re gonna remember that something interesting happens with absolute values

at that point along the number line or at some value that allows the expression

inside the absolute value to equal zero like I said absolute value 5 is positive

5 it doesn’t do anything to the left of 0 though the absolute value is going to

change the sign of that value so what makes this expression inside the

absolute value symbol equal to 0 well of course that’s going to be x

equals 2 so there’s no coincidence that we’re being asked to find the limit as X

approaches 2 something interesting is happening there really for two reasons

not only do we have an absolute value function in our expression but of course

we also have X minus 2 in the denominator and as we saw with our

earlier lessons and the other practice in your homework that you know if you’re

going to find the limit of this expression using

properties of limits you want to rewrite this expression into some other form

that allows us to plug in the value of to come up with an expression or

function that agrees with this function everywhere except x equals two well how

do we do that well just like all the other problems in

your homework we’re going to factor the numerator and maybe hopefully see it

cancel with some kind of factor in the denominator allowing us to plug in two

but at the same time let’s set up that piecewise function we’re gonna have

let’s say f of X is equal to and that bracket is going to need to be bigger

but it’s going to be equal to well let’s see let’s go to the left of two if we

plug in a value that’s to the left 2 of 2 or less than 2 like 3 3 beats with the

grape to the right how about 1 1 minus 2 is negative 1 and the absolute value of

negative 1 is positive 1 you have to change that sign so if we let the

condition be that X is less than not 0 but – we’re gonna have x squared plus 3x

minus 10 over now we’re going to take this absolute value function out and

replace it with a set of negative parenthesis so that we account for the

idea that if we plug in an x value that’s less than 2 like 1 1 minus 2 is

negative 1 we got to change the sign of that as we take the absolute value and

you know of course have it be positive 1 so we introduce the negative parenthesis

to account for what those absolute values function or notation would do but

when X is greater than 2 like 3 3 minus 2 is 1 and the absolute value of 1 is

just 1 the absolute value symbol doesn’t actually do anything so we have x

squared plus 3x minus 10 over X minus 2 well now that we have our piecewise

functions set up let’s kind of enough and notice I didn’t do like an equal

sign here because I’m doing some side work you know this if all of this right

here is not equal to this limit notation I’m just getting ready to do that

and we have a quadratic in the numerator that’s easily factorable so let’s go

ahead and do that and you can see now like we did when we were graphing

rational functions in algebra 2 or maybe even algebra 1 and certainly and our

other homework problems we’ve had so far where we’re talking about finding these

limits using properties of limits and not having to do those annoying tables

and graphs that this common factor of X minus 2 is going to cancel out of each

of these expressions and we also have a negative that’s down here in the

denominator that’s gonna you know X minus 5 divided by negative 1 is going

to be the same as negative 1 times X plus 5 then I say X plus 5 there right I

hope and let’s get this rewritten one more time and I made some graph paper so

I could show you what this graph looks like but it’s going to be getting in the

way so let’s get that out of here for now put that back up in a second we have

f of X is equal to negative X minus 5 when X is less than 2 and X plus 5 when

X is greater than 2 now of course this is negative X minus 5 because we’re

letting this negative one in the denominator X divided by negative one

and 5 divided by negative 1 is negative X minus 5 now let’s bring in that limit

notation again what’s going on here well you notice that there’s no indication of

a plus or minus in this upper right hand corner that means we’re being asked for

a two-sided limit well with piecewise functions of course when you have a

different expression to the left of 2 in to the right of 2 you can’t just plug

we’re not just going to plug 2 into some kind of expression it’s got to be it’s

going to be broken up so what I’m going to really write here now is that we’re

looking at the limit as X approaches 2 of f of X it’s a piecewise function

though and the only way that a two-sided limit exists is at the left and right

hand limits or so what we really need to do is find out

what’s the limit as X approaches 2 from the left of what I’ve renamed this

function f of X well we’re going to plug in something to the left of 2 well now

here if I do 2 minus 2 I get 0 and I can’t divide by 0 so it up didn’t like

kind of do some table again to approach from the left or right but I’ve

rewritten with the factory now I was able to rewrite this expression so we

have now just simply negative X minus 5 so when we’re approaching from the left

and I’m going to plug 2 into this expression I’m going to write 2 you know

not like 1.999 because by writing 2 or plugging or

substituting 2 into the expression defined for when X is less than 2 then

I’m taking care of or showing that I’m using that left-hand side and we get

negative 2 minus 5 is negative 7 now if I plug in or think about again

two-sided limit let’s check the left now let’s check the right well if we’re

using values to the right of 2 or greater than 2 we’re plugging it into

this other expression and we have 2 plus 5 is equal to 7

well our left and right hand limits are not equal as X approaches 2 from the

left we have negative 7 as X approaches 2 from the right we have positive 7 so

the limit as X approaches 2 of f of X that answer is going to be undefined now

if I bring back my gap my graph paper here a little bit I love that thing we

see you know if we wanted to take a look at this problem graphically when you

come back here to the original problem it looks like we have some kind of

parabola but we have a second-degree expression being divided again in two

parts because the absolute value but some kind of single degree expression so

what looks like maybe be a parabola of course is just simply of line and after

we’ve set up the piecewise function based on the break in the absolute value

and looking it to the left and the right of

two we have negative five so a y-intercept of negative five and we have

a slope of negative one so all those points represent you know some kind of

connection for that line in case you needed some help graphing piecewise

functions we have X is y is equal to X plus five so we’re looking at a liner

Sept of five one two three four five we have a slope in front of our x value of

course of one well in calculus we should know that of course and we have this

graph of the piecewise function but of course that’s not really correct we’re

just using the negative X minus 5 for X values that are less than two so as we

come through here we’re gonna come to to put an open dot because of course X

cannot be two you can’t divide by zero and to the right of two well I got this

whole line drawn but really the only part that I want is to the right of two

and here you can see one more time as we approach 2 from the left we have

negative seven and as we approach 2 from the right we have that pot that value of

positive seven and that’s our first example now our second example is going

to be a little bit harder because it’ll be very similar quadratic in the

numerator absolute value in the denominator but we’re not going to be

able to cancel out that expression that we’re getting from the denominator

what’s that going to look like so for a last example we have the limit as X

approaches 3 of x squared minus 5x plus 4 over X minus 3 now with this one we

can see the numerator is again another very easily factorable polynomial and we

have x squared minus 5x plus 4 well that’s going to factor to be let’s see

that’s going to factor to be X minus 4 times X minus 1 great but neither one of

these factors is course going to work with or handle very well that absolute

value of X minus 3 is not gonna do any cancellation so for this particular

problem we can rewrite it again into that

piecewise function like we showed in the previous example or we can understand

that really kind of like nothing very interesting is going to happen here in

terms of being able to algebraically manipulate this into a form that’s going

to allow us to use this x value of three let X approach three so we can say well

we’re going to find the limit as X approaches three from the left for this

expression and right off the top of my head my calculator you know we can say

well let’s go ahead and let X approach 3 from the left let’s go ahead and use the

value of 2.99 and when you work that out that’s going

to come out to be negative two hundred point nine nine and as you want to find

the limit as X approaches three from the right understanding that we’re going to

have to come in from you know come in two three on the from the right you know

for three and a half three point two three point one we’re going to get this and that comes out to be negative one

and the calculator is turned off negative one hundred ninety-eight point

nine nine and we can also look at this graphically but this time with there

being no way to deal with that factor in the denominator of X minus three and of

course you can’t really deal with that factor of X minus three until you

rewrite this into a piecewise function so you can allow that to actually cancel

because you can’t you can’t just cancel an expression that’s you know inside of

an absolute value function but you’re going to have a vertical asymptote at X

minus three kind of like think about with your skills and algebra two and pre

out algebra 2 and precalculus that when you

were graphing rational functions if you had a factor in the denominator that was

able to cancel you’d have a hole in the graph I didn’t have a hole in the

previous example because of you know having the extra complexity of the

absolute value function there but if you had a factor in the denominator that did

not cancel you’d end up with vertical asymptotes well you also know now that

we’re dealing with you know like trig functions that when you have a factor in

the denominator that you know that can be or avow or some kind of term that can

be zero you don’t always get vertical asymptotes like the limit as X

approaches zero of the sine of X over X you know you don’t have a the graph does

not have a vertical asymptote there at x equals zero but this one is going to

because you don’t have any fancy trig functions or anything going on there and

I don’t matter what this graph looks like off the top of my head but as you

approach 3 from the left and the right it’s going to be going down to infinity

on both sides so the limit as X approaches 3 of x squared minus 5x plus

4 over the absolute value of X minus 3 comes out to be negative infinity now

really when we deal with limits we want real values and I’ve had some some

viewers on YouTube kind of agree and disagree with with my videos that I’ve

posted for my calculus like we have a section about infinite limits and really

by me saying that the limit as X approaches 3 for this expression saying

that it’s negative infinity I am actually saying that there is no real

limit but by identifying it as negative infinity both of course approaching

negative infinity from the left and the right from this statement I can see a

behavior that I would have in the graph and that is the end of my second example

again if you need more examples or discussions lessons about limits we’ll

have a link in the description as well to my entire calculus playlist that’s

organized by chapter and topic but for now I miss a true BAM

go to your homework and if you’re still watching I just want to send out a thank

you for all the prayers and well wishes and thoughts sent our way as we were

preparing for hurricane Irma we came through just fine we were very very

blessed and we’re sending our prayers out to everyone that is rebuilding still

suffering without their you know place to live without power having struggles

with food and getting gassed and all that stuff our thoughts and prayers go

out to those and thank you so much for those you sent our way thank you you