BAM!!! Mr. Tarrou. In this lesson we are going
to learn about finding limits of function.
The same thing we learned in Calculus in chapter
one. Back in chapter one we learned how to
find limits both graphically and numerically.
Making a graph of a function and just looking
at the graph using the trace function or zoom
function in our calculator to see what that
limit. Or we found limits numerically by doing
those very tedious tables where we let x approach
from the left and the right and we noticed
when the left and right hand limits were equal
then the limit existed. Then we learned how
to solve or find limits analytically. That
involved algebraically, if needed, manipulating
the function to allow us to find let's say
like this example here a limit as x approaches
two. Well that is fine but algebraically manipulating
these expressions is not always possible and
it certainly is not always the fastest method
of finding a limit. So what we are going to
do in this video is see how l'Hopital's Rule
will allow us to find some limits we are not
able to without the aid of a table or a graph.
It will also speed up the process of finding
certain limits. And there are may different
forms of indeterminate so in this video we
are going to be properly defining l'Hopital's
Rule and working through nine examples to
help you get through your homework. So when
finding limits the answers of zero over zero
and infinity over infinity are called indeterminate.
These answers do not guarantee that a limit
exists, and if it does exist there, that there
is like this example here where x is approaching
two, we do not know what limit is. And I have
two examples here but I am going to move this
board out of the way to finish our first page
of notes so let's do this first. Find the
limit as x approaches two for this rational
function where there is a polynomial in the
numerator and denominator. For a lot of functions,
if you want to find the limit as x approaches
a particular value you can just simply plug
it in. Well if you plug two into the numerator
and denominator you are going to get an answer
of zero over zero which is an indeterminate
form. That does not mean that the limit does
not exist, not does it tell us if it does
exist what the limit is. We did these problems
earlier in the year and we don't need a graphically
representation of this function or even a
table to figure out what the limit is equal
to. I can algebraic manipulate this expression,
or this function, to allow us to find the
limit. It is going to be the limit as x approaches
two of...let's see here. We are going to have
5x and 1x and if we put a factor of seven
here and a factor of 2 here for our factors
of 14, we are going to have 5 times 2 is 10
and 7 times 1 is seven... we have a middle
term which is negative three x, so the numerator
is going to be factored to be 5x plus 7 and
x minus 2. Our denominator is going to factor
out to be x plus 3 and x minus 2. Well that
common factor of x-2 is going to cancel out
of our function. And in case you forgot, if
we were to graph this that means that this
function has a hole at the x value of 2. What
we do end up with, as far as finding limits
is the limit as x approaches 2 of 5x plus
7 over x plus 3. Now I have an expression
or function that I can plug 2 into and we
get 5 times 2 is 10 plus 7 is 17 over 2 plus
3 is 5. So we were able to algebraically manipulate
this expression so that we can get it out
of the indeterminate form and find out what
the limit is. So as x approaches two, the
limit of the function y is equal to this rational
function, is 17 over 5. There would be a hole
in this case at (2,17/5). But here we have
the limit as x approaches 0 of two times e
to the 3x minus 2 over x. Now if we let x
approach zero here, we are going to get 2
times e to the 3 times 0 power minus 2 over
0. That comes out to be 2 times anything to
the zero power is 1 so 2 times 1 is 2 minus
2 is 0 over 0. That is the same indeterminate
form I had with this problem, but there is
nothing I can do with this function y is equal
to 2 times e to the 3x minus 2 over x, to
get it out of this indeterminate form and
that is where l'Hopital's Rule is going to
help us out. So let me get this out of the
way. My high tech picture in picture. We don't
have an example anymore and this is just going
to summarize what I said in written words.
The indeterminate form cannot be evaluated
by algebraic manipulation for this example.
When transcendental, an example of that would
e to the 3x, when transcendental and algebraic
functions are combined in the same expression
this often the case. I can't manipulate this
out of, algebraically out of the indeterminate
form. Let's get the second page of notes up
so we can get l'Hopital's Rule defined and
learn when we can us it. You cannot use it
all the time and one of our examples are going
to show when you can get in trouble for just
arbitrarily using l'Hopital's Rule for every
single problem and like I said go through
those many examples in the video. Bam! Nanananana
l'Hopital's Rule. Let f and g be functions
that are differentiable on an open interval
(a,b) containing c, except at c itself. Like
my algebraic example. Functions are not differentiable
at hole. And also assume that g prime of x
is not equal to zero for all x in the open
interval from a to b, except possibly at c
itself because then you would be dividing
by zero and that is not good. If the limit
of f(x) over g(x) as x approaches c produces
the indeterminate form zero over zero then
the limit as x approaches c f(x) over g(x)
is equal to the limit as x approaches x of
f prime of c over g prime of c. Unlike in
the previous board when I did that algebraic
example I had to factor the numerator and
denominator and see if there were any common
factors that cancelled, I can just take the
derivative of the numerator and denominator
separately and then find the limit of each
and that is the limit as x approaches c of
my original expression f of x over g of x.
That is going to save us a lot of time. And
on that second example that had the transcendental
function, that is going to allow us to find
the limit period where I couldn't with algebraic
manipulation so this is very powerful. Now,
I have this in yellow, I was going to write
this in red but sometimes that does not show
up in the camera. This is very important.
If the limit of f(x) over g(x) as x approaches
c produces the indeterminate form of 0 over
0, and I will give you some alternative forms
on the other side, but if you don't meet the
conditions to apply l'Hopital's Rule and you
just apply it all the time it will give you
a wrong answer. There is a condition for using
l'Hopital's Rule, make sure that you check
it. provided the limit... I guess I could
just keep reading the rule right. Then the
limit is blah blah blah... provided the limit
on the right, that is this thing right here,
exists (or is infinite). Now an infinite limit
is not a real limit but it is still something
that we can say. Like the function approaches
infinity. If a function at c is undefined
but as approaches infinity from the left and
from the right, then the left and right hand
limits are equal thus the limit exists but
saying that the limit is infinity, or infinite,
is also saying the REAL limit does not exist.
But if approaches infinity from the left and
the right then we do have an infinite limit
just not a real limit. This result also applies,
I am putting this back into yellow again because
I want to come back to this condition over
here. This result also replies, the result
right there...l'Hopital's Rule, if the limit
of f(x) over g(x) as x approaches c produces
any one of these indeterminate forms infinity
over infinity, negative infinity over infinity,
infinity over negative infinity, or negative
infinity over negative infinity. If you try
and find a limit and you get 0 over 0 or any
of these other variations, then you can apply
l'Hopital's Rule and again how this notation
reads....You are taking the derivative of
the numerator and denominator separately,
you are not taking the derivative of the quotient.
You are not applying the quotient rule to
find the derivative. It is separate and that
is also a common mistake that students make.
I have got four examples coming up before
we start talking about alternate forms of
indeterminate that we have besides this zero
over zero pattern. That is why I ended up
with so many examples. First example coming
right up. Our first real example is getting
back to that previous example that we could
not figure out...not with algebraic manipulation
of the function, you would have to go back
to a table of values and use a numerical approach
or use a graphing utility to find the limit
of this function as x approaches zero. Remember
it was zero over zero, the indeterminate form
that says hey you can apply l'Hopital's Rule
to figure out this limit. So let's do it!
Find the derivative of the numerator and denominator
separately. We have the limit as x approaches
zero of 2 times e to the 3x, now that is not
just a simple x and the derivative of e to
the u is e to the u du or u prime, so we are
going to multiply by the derivative of 3x
which is just three. Again this is all taking
the derivative with the respect to x which
maybe I should have written since this is
the first example. Again finding that derivative
of the numerator and denominator separately.
You will probably not want to write that notation
and use an extra line of work to write that
too many times. But again, this is just to
stress that you are not taking the derivative
using the quotient rule, it is the numerator
and denominator separately. There we go. So
the derivative of e to the u is e to the u
times u prime and that is a constant so it
is zero. The derivative of x with respect
to x is simply equal to 1. Now when we go
ahead and let x approach zero we have two
times e to the zero power times three over
one. That comes out to be one times two times
three is six over one. And WOW, that did not
take a whole lot of effort. Just make sure
that we try to find the limit, we got the
indeterminate form of zero over zero, that
says that we can use l'Hopital's Rule, find
those derivative of the numerator and denominator
separately, and BOOM just like that got the
answer. Very very nice process for finding
the limit of an expression as x approaches
c. Now over here we have the limit x approaches
zero from the left. We do have a transcendental
function and some polynomials mixed in together
so let just see what happens first. We have
e to the zero power minus one minus zero over
zero to the third power and that of course
is going to be 1 minus 1 which is 0 over 0.
I am putting a check mark not because I know
what the limit is but because I don't know
what the limit is, but that indeterminate
form of zero over zero says HEY you can use
l'Hopital's Rule to finish this problem up.
So that means that we have the limit as x
approaches zero from the left, and of course
you can see that we can apply l'Hopital's
Rule to one sided limits as well, not just
to two sided limits. The derivative with respect
to x of e to the x is e to the x, the derivative
of a constant is zero, the derivative of negative
x with respect to x is equal to negative one.
In the denominator we have 3 x squared. We
applied l'Hopital's Rule, we must be ready
to go. Let x approach zero from the left,
well again e to the zero power is going to
be equal to one so we have 1 minus 1 is 0
over...uhh. 3 times 0 which is again 0. We
applied l'Hopital's Rule but what we are seeing
here is that you may not be able to apply
it just once, you may have to apply l'Hopital's
Rule multiple times. By the way, there will
be examples where you realize that I applied
l'Hopital's Rule when it was appropriate and
again I got another indeterminate from and
I applied it again and again, and you realize
you keep applying l'Hopital's Rule but you
still end up with answers that are indeterminate.
So there are cases where l'Hopital's Rule
won't help you and you will not know that
until you work out this process a few times
and it just does not seem to be working out.
We have nine examples so let's see if we happen
upon one of those examples. But let's just
apply this again. The limit as x approaches
zero from the left, again we have the condition
that allows us to apply l'Hopital's Rule,
that is a constant...that is zero, the derivative
of e to the x is e to the x again with the
respect to x. Again bringing this power down
we have 6x. Now letting x approach zero from
the left we have e to the 0 power is again1,
but here in the denominator...we don't have
a zero in the numerator so I think we are
going to have a final answer here. We are
going to let x approach zero, well 6 times
0 is equal 0 and we have the 1 over 0 answer.
As x...as we approach this 1 over 0 type answer
or scenario that will approach either positive
or negative infinity. That is were we need
to come back here and realize that we are
approaching zero from the left, so all of
our x values must be negative. Now a positive
base raised to a negative power, that is just
kind of e to the negative x is just going
to be 1 over e to the x so that is going to
be a positive answer or positive 1. But down
here, this x times a negative number, we are
going to be approaching zero but that zero
that we are approaching, all of these values
are going to be negative down here. It will
be like negative .0000000001. And then ultimately
we are looking at one over a, zero has no
sign if you will, but that denominator isn't
really zero...it is approaching zero and is
going to be a negative value. Here our limit
is going to be negative infinity. Indeed,
if you were to put this into your calculator
and graph it, it is going to be approaching
negative infinity as you approach zero from
the left and approach positive infinity as
you approach zero from the right. This is
not the entire function, it is just what is
happening as you get close to zero. So if
I had let that approach from the right, my
answer would be positive infinity and not
negative infinity. So we have got two more
examples before going on to some other types
of situations using l'Hopital's Rule. BAM!!!
For our third example we have the limit as
x approaches infinity of x squared plus x
over e to the x cubed. And letting x approach
infinity, what do we have? We have infinity
squared is a bigger infinity and another infinity
is a big ol infinity, and e a positive base
raised to infinity is going to be infinity....AND
BEYOND!!! So that means that is a form of
indeterminate that is going to allow us to
apply the l'Hopital's Rule. So let's find
the derivative of the numerator and denominator
separately. L'HOPITAL!!! The limit as x approaches
infinity of 2x plus 1 over, again finding
the derivative with respect to x, and the
derivative of e to the u is e to the u times
u prime or du. The derivative of x cubed is
3 x squared. Excellent, now let's find the
limit. The limit as x approaches infinity
of 2 times x is ....infinity plus...this is
still going to be infinity..uhgg. This is
not looking good. Now as we let x approach
infinity again, ok. So this is another one
of those problems. Now what are we looking
at in this example. Well first 0 over 0 is
not the only form of indeterminate form that
allows us to apply l'Hopital's Rule. Again
we might need to apply l'Hopital's Rule multiple
times. Let's see what happens. We have now
the limit...excuse me L'HOPITAL'S!!!... as
x approaches infinity of the derivative of
2x is equal to 2 over...uhmm. So now let's
see here. We have a variable here times a
variable here so we are going to the product
rule for derivatives. First times the derivative
of the second which is going to be times e
to the x cubed times 3 x squared, that is
first times the derivative of the second and
we are using that chain rule again, that the
derivative of e to the u is e to the u times
u prime. Now plus the second times the derivative
of the first and that is going to be 6x. Cleaning
that up we have the limit as x approaches
infinity of 2 over 3 times 3 is 9. X squared
times x squared is x to the forth. e to the
x cubed plus 6x e to the x cubed. And as you
let x approach infinity here we are going
to have 2 over...well both of these terms
have an x in them right, and that x is approaching
infinity so that is going to be some type
of infinity plus another type of infinity,
so we have 2 over infinity. If the denominator
approach 0 that means you have a large number
what is effectively over a very small number
and that means your function is going to approach
infinity or negative infinity, but if your
denominator is becoming infinity large or
becoming... compared to that numerator which
is a constant, that is going to be approaching
zero. Let me make sure I have not made any
mistakes here. Looking good. So for our forth
example we have the limit as x approaches
0 of 1 plus sine of x over cosine of x. Let's
see here. Let's just apply l'Hopital's Rule.
The limit as x approaches zero, the derivative
of 1 is 0, the derivative of sine of x is
cosine of x. And the derivative of cosine
of x is equal to negative sine of x. As you
let x, now remember x is in a trig function
so it is an angle measure right, so as you
let x approach 0 and of course are are talking
about positive rotation right, so as you let
x approach 0 the cosine of 0 is going to be
equal to 1. The sine of 0 radians is going
to be equal to zero, but there is a negative
out front. So it would appear that the limit
as x approaches 0 for this expression is going
to be negative infinity. No it is not! No
it is not..no it is not! Why? Because, just
because I am in a section about l'Hopital's
Rule doesn't mean that every single problem
is going to be worked with or be done using
l'Hopital's Rule. I never checked to make
sure that this limit was indeterminate. So
if I come back up here and I say let x equal
zero, or approach zero, well that is going
to be 1 plus the sine of zero is equal to
zero. And the cosine of zero is is equal to
one. My answer was one. My limit was not in
indeterminate form and I just arbitrarily
applies l'Hopital's Rule and I got the wrong
answer. So again, make sure you check the
condition for l'Hopital's Rule before you
just go on and apply it. Come man, CHECK YOURSELF:D
Check your problem. BAM! New type of problem
or setting. indeterminate products of 0 times
infinity. Now here we have three limits and
they all equal different answers. The limit
as x approaches 0 x times 4 over x equals
4. The limit as x approaches 0 from the left
of x squared times the cosecant of x is equal
to zero. And the limit as x approaches infinity
of x times sine of 1 over x, well that is
equal to one. But if you look at each of these
forms, let's get some red here. I think this
will be just enough to be able to read. If
you let x approach 0, you going to have 0
times and then 4 over 0 is going to be infinity.
The limit as x approaches zero from the left,
well zero squared is going to be equal to
zero and the cosecant function, that is the
reciprocal of sine. At the x y axis, the sine
is at (0,0) and then goes up to one, and it
goes in the opposite direction approaching
negative one as you approach negative pi over
2. So if you think of the sine function getting
blown inside out as the cosecant function
is, I don't know what I am trying to do with
my arms, but the cosecant function...I think
I said the wrong function there a second ago.
The cosecant function as x approaches zero
from the left is going to go to negative infinity.
And over here we have the limit as x approaches
infinity, well that is going to be infinity
and then we are going to be taking the sign
of one over infinity is basically zero and
the sine of zero is equal to again zero. So
all three of these problems are in this indeterminate
form of 0 times infinity but yet they all
give us different answers. So if the limit
as x approaches z of f of x is equal to 0
and the limit as x approaches a of g of x
is equal to infinity or negative infinity
as we have here, then we do not know the value
of the limit as x approaches a of f(x)g(x)
or even if it going to exist. All of these
three examples have real limits. If f is the
stronger function the limit will be zero,
if g is the stronger function the limit will
be either positive or negative infinity, or
there may be a finite non-zero answer...some
kind of balance between these two functions.
Like we had here. There was some kind of balance
reached between these two functions, the first
one going to 0 and the second one going to
infinity and they balanced out to be four.
So what you want to do with these problems
is you want to rewrite f times g into either
format of f over 1 over g or g over 1 over
f, to convert this 0 times infinity or infinity
times 0 limit that we have into the indeterminate
form again of 0 over 0 or infinity over infinity
to apply l'Hopital's Rule. So we are saying
HEY there is more that these two, and we had
those little sign changes there of like positive
or negative infinity, but there are other
forms of indeterminate limits, but if you
want to apply l'hopital's rule you have to
have 0 over 0 or infinity over infinity pattern
or answer in our limit so that we can apply
l'Hopital's Rule. So you know what that means,
TIME FOR MORE EXAMPLES! Hi, I am Mr. Tarrou
and I just recorded this scene twice because
I didn't realize my card was out of memory.
I just thought I forgot to hit the record
button. hehe.. Ok, so this indeterminate form
of products take three. Even though it was
right the first two times you didn't get to
see it:( As the limit as x approaches zero
of x times four over x. So we know that this
was zero times infinity, an indeterminate
form of products and yet we knew the answer
was equal to four. We can rewrite this, we
need to try and rewrite this into a quotient
which allows us to apply l'Hopital's Rule
by getting it into that indeterminate form
of 0/0 or infinity over infinity. That said
we can take the first or second factor and
flip it down. We have the limit as x approaches
zero of 4 over x divided by, again that is
going to be x over 1...that little bit I had
to erase because I just said this 5 minutes
ago:) ...we are going to take that x over
1 flip it down to make it 1 over x. Now, in
case you asking why am I allowed to just flip
that down, how does that work? Well this is
one fraction divided by another fraction,
and when you divide fractions you have to
multiply by the reciprocal of the second fraction
and that is going to be the denominator. Thus,
this would get flipped up and become a multiplication
of x over 1. Now I should be checking to see
if I can apply l'Hopital's Rule which then
I could say this is in indeterminate form
of 0/0, let's find the derivative of the numerator
and denominator individually. But, in this
particular case the x's just cancel out and
we end up trying to take the limit as x approaches
0 of a constant. I never even got to apply
l'Hopital's Rule, this is just the limit of
a constant. And of course, that is going to
be equal to 4. Over here we have the limit
as x approaches 0....ohh...yeah. You can rewrite
this so that you can have the division of
the reciprocal of either one of these factors.
Now sometimes the way you decide to set it
up is going to make the problem much easier
or hopefully not harder. So play around with
that. With this particular problem, just to
show you that it can work in both directions.
I am sure there are some examples where you
really want to rewrite it in one particular
way over the other. If I leave this x over
one in the numerator and take this 4 over
x and flip it down to be x over 4, well now
I can take my numerator and denominator an
multiply by 1 over x. That is going to cancel
the x's out and we have the limit as x approaches
0 of 1 over 1/4. Dividing by 1/4 is the same
as multiplying by 4 and so we still end up
of course trying to take the limit as x approaches
0 of 4! Which is still going go be 4. Alright,
next example. Now I really probably don't
want to take the derivative of cosecant if
I don't really have to so I am going to rewrite
this so I am going to divide by the reciprocal
of my first factor. We have the limit as x
approaches 0... We already discussed on the
previous screen how all of these are the indeterminate
form for products or multiplication. So as
x approaches zero of... Excuse me. I am going
to leave, I said that wrong, I really don't
want to take the derivative of cosecant if
I don't have to so I am going to leave x squared
on time (I just said that backwards) and bring
the cosecant down into the denominator as
1 over cosecant of x. Now of course I have
cosecant of x over one, flipped it down dividing
by the reciprocal. But 1 over cosecant of
x we know is sine of x. And now let's make
sure that we can apply l'Hopital's Rule before
we just start doing so without thinking. We
are letting x approach zero so we have zero
squared which is equal to zero and the sine
of zero is also equal to zero. So that is
the indeterminate form now that will allow
us to apply l'Hopital's Rule. So finding the
derivative of the numerator with respect to
x, that is going to be 2x. The derivative
of sine is cosine. So the limit as x approaches
0 of 2x over cosine of x, and as we let x
approach 0 now we are going to get 2 times
0 over the cosine of 0 which is equal to 1
and that is equal to 0. The answer I said
this limit is equal to in the previous screen.
Now the last one. The limit as x approaches
infinity of x times the sine of 1 over x.
Well again if I bring the sine of 1/x down
and divide by the reciprocal, that 1 over
sine is going to become cosecant which may
be fine..may work just fine but why introduce
a harder trig function when I don't really
need to. So, we are going to divide by the
reciprocal of the first factor... or rewrite
this as the limit as x approaches infinity
of sin of 1 over x over 1 over x. Trying to
find the limit at this point I have the sine
of 1 over infinity is 0, so the sine of 0
is again equal to 0. This is equal to zero
in the denominator. So we are good to finish
up by applying the l'Hopital's. So the limit
as x approaches infinity. The derivative of
sine is cosine of 1 over x, now we need to
multiply by the derivative of that inside
function. Now 1 over x is the same as x to
the negative 1 power. The derivative of...let's
see. I am going to take that negative 1 and
bring that power down out front and we are
going to have negative one and we are going
to reduce that power by one giving us negative
x to the negative two power...or negative
1 over x squared. Down here we have the same
thing. The derivative of 1 over x is going
to be negative x to the negative two power.
These negatives are going to cancel out, the
x to the negative two's are going to cancel
out, and we are left with the limit as x approaches
infinity of the cosine of 1 over x. Letting
x approach infinity, we are going to have
1 over infinity is going to be zero. The cosine
of zero is equal to 1. That is end of these
three examples all involving indeterminate
form for multiplication or products. Now we
are going to move on to the indeterminate
form of power. Before we move on to the indeterminate
form of power, I just realized just before
I was about to erase the board that in my
frustration of having to shoot this three
times when I didn't make a mistake except
for not realizing that my memory card was
getting close to full, to include the notation
that we were approaching zero from the left.
Now moving on to POWER! Nananana... Indeterminate
Powers. If the limit as x approaches a of
f of x raised to the g of x power...so you
are looking at a function basically or an
expression raised to another expression...yields
either zero to zero power, or 1 to the infinity
power, or infinity to the zero power, then
you can start trying to get that problem into
a form that allows you to apply l'Hopital's
Rule by taking the natural log of both sides
of the function. That is going to allow us
to then...so we have.... We are basically
going to say lets let y equal f(x) raised
to the g(x) power. We are going to take the
natural log of both sides. The power rule
for logarithms is going to allow us to take
that variable exponent and move it down out
front as a multiplication. Then to get the
natural log function away from the y, we are
going make both sides of the equation an exponent
of e...or just write into exponential form.
That gives us y is equal to e raised to g(x)
times the natural log of f(x) power. Then
kind of see what happens after that. This
next example is going to be relatively long
so I need to erase this board so I have enough
room. Let's find the limit as x approaches
0 from the right of 5 times x raised to the
x over 3 power. You look at this the way it
sits right now, we are going to let x approach
0 from the right and x is going to become
infinity small but always positive. That means
we are going to have five times 0 which is
0, raised to 0 over 3...again a positive 0
approaching 0 but positive 0 if you will over
3 which is going to be 0. So we do have an
indeterminate form for power. So I am going
to go ahead and say let's let y equal the
limit as x approaches 0 from the right of
5 x to the x over 3 power. Now I am going
to take that 5, because if I am taking the
limit of something...a function..an expression..with
a constant in front of it, I can pull that
out front and write y is equal to 5 times
the limit as x approaches 0 from the right
of x raised to x over 3 power. I want to apply
the natural log function to both sides but
I would like to get the five over to the other
side first just out of preference...just because
I think that is going to be the easiest. We
have y over 5 is equal to the limit as x approaches
0 from the right of x to the x over 3 power.
Take the natural log of both sides and we
have the natural log of y over 5 is equal
to the natural log of the limit as x approaches
0 from the right of x to the x/3 power. Now
because the natural log function is continuous
on all values where x is positive and again
we are not letting x equal zero...we are letting
x approach 0 from the right. This is a continuous
function, so I can take this natural log function
and bring it into the limit notation and have
the natural log of y over 5 is equal to the
limit as x approaches 0 from the right of
the natural log of x to the x/3 power. Now...
fix that parenthesis there. Normally when
I say that I am going to apply the natural
log function, or the log of any base to a
variable I wrap that inside an absolute value
function because you cannot take the log of
a negative number. But again we are letting
x approach 0 from the right so x is always
going to be positive and thus the absolute
value symbols are not necessary. Now we are
going to take this out of this exponential
form so we can write this...I don't know if
it is going to be 0 over 0 or infinity over
infinity or what...but to get into a form
with which we can apply l'Hopital's Rule again
that 0 over 0 idea, we are going to take the
x over 3 power and move it out front. We get
the natural log of y over 5 is equal to the
limit as x approaches 0 from the right of
x over 3 times the natural log of x. Now if
you look at what we have going on here. If
we try to apply the limit here this is going
to be zero and this is going to be negative
infinity, so it is still in indeterminate
form but no a form where we can apply l'Hopital's
Rule. So we are going to....let's see... How
about... I think it is going to be a little
bit difficult to write a 1 over the natural
log of x type of situation so let's rewrite
this multiplication of x over 3 as a division
of it's reciprocal. We have the natural log
of y over 5 is equal to the limit as x approaches
0 from the right of the natural log of x over
3 over x. Now as we let x approach 0 from
the right, remembering that y equals the natural
log of x, that graph looks something like
this. As you approach 0 from the right, that
is going to go to negative infinity. As you
approach 0 from the right that is going to
go to positive infinity which is an indeterminate
form where we can use L'HOPITAL'S RULE:D So
come on Mr. l'Hopital. We are going to take
the natural log of y over 5 is equal to the
limit as x approaches 0 from the right...so
exciting...the derivative of the natural log
of x is equal to 1 over x. The derivative
of 3 over x, that is 3 times x to the negative
one. Using the power rule we are going to
have negative three times x to the negative
two power, so it is negative 3 over x squared.
Ok...So. Now if I multiply the numerator and
denominator by x over 1 which is effectively
just 1 right. These x's are going to cancel
out and that x is going to cancel out with
one of those. We get the natural log of y
over 5 is equal to the limit as x approaches
0 from the right of 1 over negative 3 over
x. That becomes, well if you take the denominator
and you flip it up that is going to become
x over negative three right. I am running
out of room so I am going to erase this and
rewrite that 1 over negative three over x,
as x over negative three. As I let my x approach
0 from the right we get 0. Now we have the
limit as x approaches 0 from the right of
x over negative three becomes 0...Now I have
a format where I can rewrite this back into
exponential form. I can let both sides be
an exponent of e and get y over 5 is equal
to e to the 0 power which is one. Multiply
both sides by five and y is equal to 5. Y
was the limit as x approaches 0 from the right
of 5x raised to x over three power. Let me
just double check my notes, we are good!!!
Now I have got one last form. It is going
to be the indeterminate form for differences
and this lesson will be done with one more
example. Thank you for watching!!! I love
this stuff:D Last form, Indeterminate form
for Differences. If the limit as x approaches
a of f(x) minus g(x) equals infinity minus
infinity, this is considered indeterminate.
If limit as x approaches a of f(x) which is
infinity, if that wins the answer is going
to be infinity. You are going to be taking
an infinitely large number and subtracting
with something that is also infinity but not
growing as fast and ultimately that is going
to go to infinity. If the limit as x approaches
a of g(x) which is infinity, if that wins,
if that is the stronger or larger infinity,
then the answer is going to be negative infinity.
Or maybe there is a balance somewhere in-between
and there is a finite limit, you know a numerical
value. Try to work these out by bring f(x)
and g(x) together to form some type of quotient
so that when you get into that quotient form
and you try and apply the limit you get a
0 over 0 or infinity over infinity or some
form of that with some signs which will allow
you to apply the l'Hopital's Rule. So our
last example finally. The limit as x approaches
zero of the cosecant of x minus the cotangent
of x. Now this time i didn't put a little
plus or minus up there on the exponent part
to indicate a left or right limit like I forgot
to do on the previous example, but this time
it was not by accident. If you approach zero
from the right, then your cosecant, you can
see we have one period graphed of our cosecant
of x. I hope you can see the purple lines,
two period actually because the period of
cotangent is pi where the period of cosecant,
secant, sine, and cosine are all 2pi. At any
rate. As I approach 0 from the right both
of these functions are going to positive infinity.
So this is going to be infinity minus infinity
which is our indeterminate form for differences.
If you let x approach zero from the left,
I want to keep saying negative, but as you
approach zero from the left both of the functions
are going down to negative infinity. Now that
would be negative infinity minus negative
infinity which will is negative infinity plus
infinity which is still infinity minus infinity.
So either way you look at this, left or right
limit, this is going to have an indeterminate
form of infinity minus infinity. So let's
see if we can use this tip here of how to
work out these problems and write this in
terms of a quotient. Well cosecant of x minus
cotangent of x, nothing is going to happen
there. There is nothing to do, it is just
two terms with no common denominators. Oh
yeah that is right, but if I rewrite these
functions like I did doing trig proofs in
Precalculus or Trigonometry, if I rewrite
this in terms of the most basic trig functions
sine and cosine I think we are going to have
some fractions that we can then use to find
common denominators with and combine those
terms. So we have the limit... And again this
is infinity minus infinity. We have the limit
as x approaches zero of cosecant which is
1 over sine of x minute cotangent which is
cosine of x over sine of x. I don't even have
to find common denominators, already have
them there. So let's write that as one quotient.
We have the limit as x approaches zero of
1 minus cosine of x over sine of x. Now we
have just have a quotient. Maybe we get the
answer now because earlier in the video I
did a problem like this and I did l'Hopital's
Rule immediately and did that when I should
not have because we could find the limit.
l'Hopital's Rule gave me the wrong answer.
So as we let x approach 0, we have 1 minus
the cosine of 0 is 1, so we have 1 minus 1
which is zero. The sine of zero is zero. Ok,
there we go, we have the indeterminate form
of 0 over 0, so we can apply l'Hopital's Rule.
We have the limit as x approaches zero of
the derivative of 1 is 0, the derivative of
cosine is negative sine but there is a negative
there so it is negative negative sine of x,
which of course those are going to cancel
out leaving me with just sine of x on top
or the numerator. The derivative of sine of
x is cosine of x. I could let x approach zero
right now, but just to rewrite this, the sine
over cosine is tangent. The limit as x approaches
zero of the tangent of x. If you remember
what the tangent function looks like, it comes
through something like this and passes through
the origin and goes to infinity as you approach
pi over 2 from the left and approaches negative
infinity as you approach negative pi over
2 from the right. I don't care about approaching
pi over 2 or any form of it, I just want to
approach zero. The tangent of zero is equal
to zero. That is the end of my last example.
I am Mr. Tarrou. BAM!!! GO DO YOUR HOMEWORK:D
