BAM! Mr. Tarrou Well, when you were in your
Trigonometry or your PreCalculus class I hope
you did not think that you would never see
those trig functions again. Sine, Cosine,
Tangent, all that good stuff. Because they
are going to keep coming back over and over
again in Calculus. So we talked about finding
limits graphically, numerically, using the
properties of limits. But what we have not
talked about yet is finding limits of trigonometric
functions. So let 'c' be any real number in
the domain of the given trig function. Let
me just step out here a little bit while I
am talking so you can copy the bottom. We
have the limit as x approaches c of sine of
x is equal to the sine of c. We have the limit
as x approaches c of tangent of x equals tangent
of c. And so on and so on and so on. Basically
what you are looking at with all of these
six trig functions is that you are going to
be able to just simply take the value of c
that you are allowing x to approach and just
plug it into the trig function. The reason
why you can do this is, and why it needs to
say let c be a real number in the domain of
the given trig function,... you know... Like
tangent is undefined at pi/2 and then plus
or minus n*pi. Of course until you mess around
with the period and everything. But tangent,
secant, cosecant, cotangent, they all have
vertical asymptotes and thus they all have
values at which they are undefined. But, within
their domain they are smooth and continuous
curves. And smooth and continuous curves pretty
much, like your polynomials, as we started
to just introduce the properties of limits...
curves that are smooth and continuous.. there
are no sharp bend, the are no holes, there
are no vertical asymptotes... Within the domain
of those smooth and continuous curves, finding
the limit is really not much more than just
taking the value of c that x is approaching
and plugging it into the function. Now there
is a lot to, you know, there is a lot of complexity
with the idea of limits but with these smooth
and continuous curves.. within their domain..
to find the limit at c we are looking at just
simply just plug in those values. We are going
to go to the next screen. There are two special
cases which I am going to back up with a couple
of graphs and then do a bunch of examples.
BAM! Here are our two special cases. We have
the limit as x approaches zero of sin(x)/x
equals 1. Now the function... We interrupt
this video to inform you that Mr. Tarrou cannot
copy his notes. I do pre-work all of my problems
out before I make these video and even so
occasionally I mess up. So I had wrote this
as 1-cos(x) over cos(x) and because I can't
talk and think at the same time sometimes
I never noticed I made a mistake in this section
of the video. So let's try that again. And
let's get two percent.. well actually let's
make sure I am right all of the time since
these are videos and you are trying to learn
from them... and there are no mistakes. Our
two special trigonometric limits are the limit
as x approaches zero of sin(x)/x is equal
to 1. If we were to graph this with our graphing
calculator and it would be easier if it is
radian mode. It will fit in the screen, your
graphing calculator screen a little better.
Put that into radians and graph y equals sin(x)
over x and you will have a graph that oscillates
along the x axis but as it gets closer and
closer to the x value of zero, which is where
we are allowing our limit to approach, it
starts to grow in oscillations and... it gets
to the biggest oscillation where it reaches
a y value, or almost reaches a y value of
1. And we have this x axis acting as a asymptote,
except the graph is just oscillating and every
time just a little bit shallower and shallower
as you go off to infinity and negative infinity.
Though of course never, you know, becoming
a straight line. It will continue to oscillate
around that x axis. So what is happening?
Well as you approach zero from the left, and
if I were to do a one sided limit it would
have a little negative up here as the exponent,
when you approach zero from the left the graph
is approaching a y value of 1. And if you
approach it from the right it is approaching
that same y value of 1. So the left and the
right hand limits are equal and our approaching
the value of one. We have the hole here, the
removable discontinuity at (0,1) because you
cannot divide by zero. Most of the time when
your denominator is equal to zero and it does
not factor out you will get vertical asymptotes.
But, indeed in this case we have a hole at
the point (0,1). So we would call that a removable
discontinuity. Again, the left and the right
hand limits are equal to one. So let's just
keep that in mind as we work through our examples
after this correction. The limit as x approaches
zero of 1-cos(x) over x is equal to zero.
Again it is going to oscillate a little bit
differently around the x axis. It is going
to bounce up and just sort of slowly start
to decrease the heights of these oscillations
and just touch the x axis as x goes to negative
infinity and positive infinity. You see over
here it is above the x axis and to the left
is below the x axis. Just touching... basically
your period of cosine. At (0,0) the graph
is coming along.. it is touching, it is dipping,
and it goes through (0,0). Well at (0,0) again
you cannot divide by zero. Again with this
particular ratio we do not get a vertical
asymptote at x=0. We have hole. So again we
a removable discontinuity. As we approach
zero from the left, with a little negative
exponent... as we approach zero from the right,
if you are writing your one sided limits that
would be a plus up here in the exponent, we
are approaching zero from both the left and
the right hand side. Thus the limit of 1-cos(x)
over x as x approaches zero is 0 with that
hole right there. And we can see from the
graph as well just to kind of bring in other
concepts before I let you get back to our
regular video, this is an even function. If
we take this graph and fold over the line
x=0 or the y axis it will perfectly line up
on itself. Where you plug in opposite numbers
and you get the same answers. F(x)=F(-x) This
is an even function. And over here, we take
this function and were to rotate it 180 degrees..
except for the labels on our x axis would
be upside down.. the graph would line up with
itself exactly. You plug in opposite numbers
and you get opposite answers. That is what
is going on with your odd functions F(x) is
equal to the opposite... or -F(x).. the opposite
of F(x). So, even functions have symmetry
to the y axis. Odd functions are symmetric
to the origin. And these two special cases
of trigonometric limits that we are going
to use as we get through our next believe
it was six examples that I did in this video.
So, I am Mr. Tarrou. Back to our regular program.
And let's see how they are used in I think
I am going to do about six or seven examples.
We will call this video done. For our first
two examples we have the limit as x approaches
pi/2 of the cotangent of x. Well the cotangent
of pi/2. Let's see. We just had that chart
up there. It said if I was finding the limit
as x approaches c of cotangent of x, that
it would just be simply equal to the cotangent
of c if this value is within the domain. So
this is going to become the cotangent of pi/2.
Ok, on the unit circle pi/2 is 90 degrees.
So... here is zero, pi/2 is 90 degrees. Now
cotangent of theta is equal to what? x/y.
So at pi/2 or 90 degrees the x value is zero
and the y value is 1. And 0 divided by 1 is
zero. So there is our answer. The limit is
equal to zero. This one though, I am kind
of throwing this in to.. What would happen
if you were not really thinking about the
domain of your trig functions and you were
just boom... boom... boom.. plugging in this
numbers. Well, the limit as x approaches pi/2
of the tangent of x, your issues with domain
are going to come from the quadrantal angles...
well I guess I say that, but of course if
there is some kind of multiple of x.. so that
would change the period. So maybe just forget
what I said. Just plug these in and see what
happens. The tangent of x as x approaches...
the limit as x approaches pi/2, well just
plugging these in, right. Or attempting to.
That is going to be the tangent of pi/2. Now
pi/2, tangent is y/x. So y at 90 degrees is
1 and the x value is 0. Well, 1divided by
0 is undefined. Well when we found limits
using the properties of limits if we approached
a value that made our function undefined,
we would find the left and right hand limits
and see if they were equal. Well actually
with algebra a lot of time we would use some
manipulation to see if we could just change
that denominator.. or cancel the factor out.
But if the left and right hand limit is equal,
then we can say that is the limit. Now if
you understand what tangent looks like as
a graph, you will understand that the left
and the right hand limits are not going to
be equal. But, without drawing a picture how
would you check for the left and right hand
limit. Well let's see. We are going to say...
let's let the limit as... That is nice handwriting.
Let's let the limit as x approaches pi/2 from
the left of the tangent of x. Again I am going
to try and explain this with out looking at
the picture, which you certainly should know
what the parent function looks like. You should
know what y=tan(x) looks like. I will draw
it for you in a second. But as you approach
pi/2 from the left, x is less than pi/2. X
is next to the trig function so it has to
be an angle measure. So our angle measure
is a little bit less than pi/2. Well if I
am on the unit circle and not looking at the
graph, less than pi/2 is between 0 and pi/2
so we are in quadrant 1. Now when we are in
quadrant 1 tangent is positive. And tangent
again is y/x, and we are in quadrant 1. The
y value is approaching 1, the x value is becoming
infinity small. So a very very small value
in the denominator so this is shooting towards
positive infinity. Whereas if I approach pi/2
from the right... The limit as x approaches
pi/2 from the right of the tangent of x. Again
out of a... You put an angle measure into
a trig function and you get out the sides
of a triangle basically, the ratios. Again
we are looking for y/x. As we approach pi/2
from the right, that means that my values
for x are a little bit larger than pi/2, so
that means if I am more than 90 degrees or
more than pi/2 radians I am quadrant 2. Well
in quadrant 2 my x's are negative and my y's
are positive. So when I do that division I
am going to get a negative answer. So indeed
when x is greater than or approaching pi/2
from the right, and I am making my hand motion
as if I am on the unit circle, if my angle
measure is just a little bit bigger then pi/2
and I am quadrant 2, then tangent is negative
in quadrant 2. So this is going to be negative
infinity. And my left and right hand limits
are not equal, thus my two sides limit does
not exist. That is why in the previous screen
I said as long as c is within the domain of
the trig function then all you have to do
is plug in c. Well pi/2 is not in the domain
of tangent of x, so that is why we are not
getting an answer. By the way just to give
you a quick quick quick sketch this is y=tan(x).
Here is our first asymptote at pi/2. Here
as approach pi/2 from the left it is shooting
up to positive infinity. Then the second period
is going to come up from negative infinity
and continue on. So as I approach pi/2 from
the left is shoots up to positive infinity.
As I approach pi/2 from the right it is going
down. So we have 5 more examples, all of those
unlike the second will have an answer. Alright.
Let's get going. We have got the limit as
x approaches 2 of the cosine of pi/3 x so
we should be able to take the 2 and plug it
in. So that becomes the cosine of 2pi/3 x.
Did you forget that unit circle from precalculus?
I hope not. I know my student's haven't. I
quizzed them on it to death. The cosine of
2pi/3. In case you are little rusty on the
unit circle, pi/3 radians is equivalent to
60 degrees. So just in case you need that.
So let's see here. One pi/3 which is 60 degrees.
2pi/3 which is 120 degrees which lands you
in quadrant 2. Not that you should have to
mentally flip back and forth between degrees
and radians that much. You really need to
know that unit circle. The cosine of 2pi/3
is negative 1/2. Next example. We have the
limit as x approaches zero of sin(x) over
4x. This looks familiar or similar to one
of those special trig functions that I brought
up. But it has a 4x in the denominator instead
of just x. So as you get more comfortable
with limits you might not need to show this
work, but we are just learning this. So I
have to break it down. [singing] break it
down. It becomes the limit as x approaches
0 of 1/4, I am going to pull that denominator
and make it a separate factor, times sin(x)/x.
That since we are just learning all of these
properties of limits is equal to the limit
as x approaches zero of 1/4 times the limit
as x approaches 0 of sin(x)/x. Well, when
you take the limit of a constant... when you
take the limit of a constant, regardless of
what the value of c is the answer is going
to be the constant itself. So the limit as
x approaches 0 of 1/4 is 1/4. And the limit
as x approaches zero of sin(x)/x we just saw
the graph where it had a hole at the y value
1. It was coming up and approaching the value
of 1 from both sides, so the limit here is
equal to 1. That final answer is 1/4. The
limit as x approaches zero of sin(x) times
1-cos(x) over x. That does not look like what
I had up here in the previous two introductory
scenes either. So I am going to something
similar to what I did with number four. That
is I have two factors in the numerator, sin(x)
times 1-cos(x). I am going to pull that multiplication
apart. In the denominator I just have one
x. Now it could be x squared and both factors
would have an x in the denominator, but I
only have one. So this is going to be expanded
to look like this. The limit as x approaches
zero of sin(x)/1. That is really unnecessary,
I am doing that for teaching... Times... the
over one part...times 1-cos(x)/x. Now I have
just made this look different. I did not change
its value. Let's see here. Sin(x) times 1-cos(x)
is what I have in my numerator. And 1 times
x is just the x. So I can now take this product
and find the limit of each factor within that
product and write the limit as x approaches
zero of sin(x) times the limit as x approaches
zero... I am running out of room here... of
1-cos(x) over x. Ok. I did not run out of
room. Ok. Limit as x approaches zero of sin(x).
[pop] zero radians, sin(x), sine is y over
r on the unit circle or just the y coordinate.
So it is 0, the sine of zero is 0... times
the limit as x approaches 0 of 1-cos(x) over
x is as we just saw in the previous introductory
scene 0 itself. And 0 times 0 is equal to
zero. BAM! Two more questions or examples.
The limit as x approaches zero of sin(x) squared,
or sine squared x, times csc(x) over x. And
you can't plug in zero because it makes the
denominator undefined. That does not look
super easy. But if you remember your trig
functions, if you remember that sine and cosecant
are reciprocals. On the unit circle sine of
theta is y/r and cosecant is r/y. They cancel
each other out. They are reciprocals. So I
can rewrite this as, or just take out just
for matter of space, I can rewrite the cosecant
of x as 1 over sin(x). Then sine squared over
1 times 1 over sine x, those are going to
cancel out. This sine in the denominator from
cosecant, the sine in the denominator will
cancel with one of the two sines that are
in the numerator. We end up with the limit
as x approaches zero of sine of x over x.
We now know that it is equal to 1. Next example.
The limit as x approaches zero of the sine
of 5x over 2x. Now this looks very similar
to the limit as x approaches zero of.. well
I have it over here... I have it over here,
so why write it again. Looks like this. Sine
of x over x is equal to 1. But this is not
an exact match to this. Now I don't really
care about the fact that there is a 5 or a
2. What I really care about are these values
are not a perfect match. Ok. This is the sine
of.. This is really the limit as x approaches
zero of the sine of something over something
equals one. What is important is those somethings
are a perfect match. You know, that it is
x and x... it is not x and y. It is not a
and b. It is the sine of x over x. This the
same variable, so thus they must represent
the same value. So what we are going to get
here in a minute is... I cannot change this
5x because the sine of 5x... That 5x is within
the sine function. So I can't really change
it. But what I can have more flexibility over
is with the x. And if I can get the sine of
5x over 5x. If I can get the limit as x approaches
zero of the sine of 5x over 5x, then that
will be equal to 1 just like this is equal
to 1. So what I am going to do is I am going
to expand this out and get the limit as x
approaches zero of.. I will take that 2 out
of the denominator and it is 2 times the...
you know, the denominator is 2 times x. And
it is a factor so I can pull it apart and
get 1/2 times the sine of 5x over x. Ok. Well,
what do you do. I still don't have a match.
This is still 5x and this is still just 1x.
Again this 5x is in the math function. It
is not sine times 5 times x. It is the sine
of 5x. So that 5 is going nowhere, not without
some identity work or something. So forget
it, it is not going anywhere. I pulled out
the 1/2 out, I pulled the 2 out of the denominator
and wrote it as 1/2. Now if I can do some
other kind of like... you know.. manipulation
with that multiplication going on there, I
will be happy. Like if I could get that denominator
to be 5x, I could be done with this problem.
So if I could just make it the limit as x
approaches zero of something over 2... I am
going to add something in... times the sine
of 5x over 5x. If I could just get a 5x to
be in the denominator. Well, I just took that
denominator and multiplied the denominator
by 5.. or I took the sine of 5x over x and
divided it by 5. So if I took this and divided
it by 5, then I could just simply cancel that
out by multiplying by 5. So what I just did
there was.. not sure if maybe you are totally
happy with that. What I kind of did here was
I came in and I multiplied that second fraction
by 5/5. Then with the commutative property
I moved this 5 over here to be on top of the
2. When I multiplied the top and bottom by
five, 5/5 is one, so I am just making it look
different. I am multiplying by one, I am not
changing its value. Well now this becomes
the limit as x approaches zero of 5/2 times
the limit as x approaches zero of the sine
of 5x over 5x. This is a constant so this
is 5/2. That is equal to 1. So that is equal
to 5/2. Now please don't be tempted even though
this pattern has shown up. You could just
be going like... there is a 5/2 and that is
your answer. Is that always going to happen?
It might. But I would strongly suggest not
trying to glance at this and seeing some kind
of short cut and thinking that is always going
to work. Because many many many many years
ago when I first took Calculus 1, and I was
just out of high school, I was working full
time, I had a girl friend who is now my wife...
But you know, a lots of distractions, so I
did not do so quite so good the first time
around in Calculus. And that was because I
was looking for these short cuts, I was not
studying enough, I was working too much, I
was tired. So I went into this Calc test,
it was integration which we are far away from
as far as us explaining it, and I got out
of that test and thought that was soon easy.
And yes, it was because I bombed it. [laughing]
You need to get these equations that you are
dealing with... When you are going to use
some kind of identity or well let's call it
an identity and pull it in and us it with
your math problem, you need to see an exact
match to make that applicable. So looking
for shortcut patterns is going to run into
big big trouble. Get that algebraic manipulation
done if you are going to use some kind of
identity or property make sure that it is
an exact match before you try and apply it.
I explained why the 5x was there and how it
could be ok. Hopefully you understood it.
This is my last example for finding limits
with Trig functions. I am Mr. Tarrou. BAM!
Now Go Do Your Work. Go Do Your Homework:)
