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