Limit examples (Part - 3) - Mathematics, Engineering Video Lecture - Engineering Mathematics

Let's do some more
limit examples.
So let's get another problem.
If I had the limit as x
approaches 3 of, let's say,
x squared minus 6x plus 9
over x squared minus 9.
So the first thing I like to do
whenever I see any of these
limits problems is just
substitute the number in and
see if I get something
that makes sense, and
then we'd be done.
Well, usually we'd be done.
I don't want to make these
sweeping statements.
If the function is
continuous, we'd be done.
But if we put the 3 in the
numerator, we get 3 squared,
which is 9, minus 18 plus 9.
So that equals 0.
And the denominator also--
let's see, 3 squared minus
9, that also equals 0.
So we don't like having 0/0.
My pen tool is
malfunctioning again.
So we don't like getting 0, 0,
0, so is there any way we can
simplify this expression to
maybe get it to an expression
that, when we evaluate it at x
equals 3, we actually get
something that makes sense?
Well, whenever I see two of
these polynomials here, and
they look, just by inspecting
them, relatively easy to
factor, I like to factor them
out because maybe there's the
same factor in the numerator
and the denominator, and
then we can simplify it.
So let's say that this is the
same thing as-- that looks
like it's x plus 3--
no, no, no, x minus 3.
This is x minus 3.
It actually looks like it's x
minus 3 squared, but we're
just going to write x minus 3
times x minus 3, which is, of
course, x minus 3 squared.
And then in the denominator,
you know how to factor these,
this is x plus 3 times
x minus 3, all right?
So the limit as x approaches 3
of this expression is the same
thing as the limit as
x approaches 3 of
this expression.
And, of course, there's nothing
we can do to change the fact
that this function, or this
expression, is undefined
at x equals 3.
But if we can simplify
it, we can figure out
what it approaches.
Well, if we assume that x is
any number but 3, we can cross
out these two terms because
then they wouldn't be 0, right?
It only is 0 when x is equal to
3 because-- so in the numerator
and the denominator, we
can cross this out.
And we can say-- and I'm not
being very rigorous here, but
this is kind of how it's
taught, and I think you get the
intuition-- that this is the
same thing as the limit as x
approaches 3 of x minus
3 over x plus 3.
Now let's just try to stick
the x in and see what we get.
Well, in the numerator,
we get 3 minus 3.
We still get 0.
But in the denominator
here, we get 6, right?
3 plus 3 is 6.
So now we get a good number.
0 or 6, well, that's a
real number, so it's 0.
0/6 is 0.
So that was interesting.
The first time we did it,
we got the answer 0/0.
And now we get the answer
0 by simplifying.
But, of course, it's very
important to remember that
this expression is not
defined at x equals 3.
It's defined everywhere but,
but if we were to graph it, and
I encourage you to do so, you
would see that as you get
closer and closer to x
equals 3, the value of this
expression will equal 0.
And I know what
you're thinking.
Well, this was 0/0.
Is every time I get 0/0 going
to end up just becoming 0 when
I evaluate the expression?
Well, let's explore that.
Let me clear this.
Let's say what is-- pen is not
working-- the limit as x
approaches 1 of x squared
minus x minus 2.
No, let's say x squared
plus x minus 2.
As you can see, I do all
this in my head, and
I'm prone to mistakes.
And all of that over x minus 1.
Well, once again, if we just
evaluate it, let's see what
happens when x equals 1.
You get 1 squared plus
1, so it's 2 minus 2.
You get 0/0.
So once again, we get 0/0, and
we have to do something to
this maybe to simplify it.
Well, let's factor the top.
So that's the same thing as
the limit as x approaches 1.
Well, that's x minus 1
times x plus 2, right?
And I think you'll often
discover when you see a lot of
limit problems that even if
this top factor, if this top
expression, is hard to factor,
chances are, one of the things
in the denominator that are
making this expression
undefined is probably
a factor up here.
So sometimes you might get a
more complex thing that isn't
as easy to factor as this, but
a good starting point is to
guess that one of the factors
is going to be in the bottom
expression because that's kind
of the trick of these problems,
to just simplify
the expression.
So once again, if we assume
that x does not equal 1, and
this expression would not be
0 and this would not be 0,
then these two could
be canceled out.
And we get that this is just
the same thing as the limit as
x approaches 1 of x plus 2.
Well, now this is pretty easy.
What's the limit as x
approaches 1 of x plus 2?
Well, you just stick 1 in
there, and you get 3.
So it's interesting.
When we just tried to
evaluate the expression at
x equals 1, we got 0/0.
And in the previous example, we
saw that it evaluated out when
you simplified it to 0, and in
this example, it came out to 3.
And I really encourage you, if
you have a graphing calculator,
graph these functions that
we're doing and see and show
yourself visually that it's
true, that the limit as you
approach, say, x equals 1
actually does approach the
limits that were solving for.
And make up your own problems.
Hell, that's what I'm doing.
So you could prove
it to yourself.
So let's do another.
Let's do one that I think
is pretty interesting.
Let's say what's the limit
as x approaches infinity?
The limit as x approaches
infinity of, let's say, x
squared plus 3 over
x to the third.
So the way I think about these
problems as they approach
infinity, just think about what
happens when you get
really, really, really
large values of x.
And kind of a cheating way of
doing this is, if you have a
calculator, even if you don't
have a calculator, put
in huge numbers here.
See what happens when x is a
million, see what happens when
x is a billion, see what
happens when x is a trillion,
and I think you'll
get the point.
You'll see what-- if there
is a limit here, you'll
see what it's going to.
But the way I think about it
is, in the numerator, kind of
the fastest-growing term here
is the x squared term, right?
This is the fastest-growing
term here.
In the denominator, what's
the fastest-growing term?
Well, in the denominator,
the fastest-growing term
is this x to the third.
Well, what's going to
grow faster, x to the
third or x squared?
Well, yeah, x to the third's
going to grow a lot
faster than x squared.
So this denominator, as you get
larger and larger and larger
values of x, is going to grow a
lot faster than that numerator.
So you could imagine if the
denominator's growing much,
much, much faster than the
numerator, as you get larger
and larger numbers, you're
going to get a smaller and
smaller and smaller
fraction, right?
It's going to approach 0.
And so as you go to
infinity, it approaches 0.
I know that I kind of just
hand waved, but that's really
how you think about it.
Another way you could do
it is you could actually
divide this fraction.
You could actually divide this
rational expression, and you'll
get something like 1/x plus
something, something,
something, and then you'd also
see, oh, well, the limit as x
approaches infinity
of 1/x is also 0.
Let's do one more.
I'll do this fast so
I can confuse you.
The limit as x approaches
infinity of 3x squared plus
x over 4x squared minus 5.
These problems kind of look
confusing sometimes, but
they're really easy.
You just have to think about
what happens as you get
really large values of x.
Well, as you get really large
values of x, these small terms,
these ones that don't grow as
fast as these large terms,
kind of don't matter anymore,
right, because you're getting
really large values of x.
And this case, these don't
matter anymore, and then
these two x terms grow
at the same pace, right?
And they'll always be
kind of growing in
this ratio of 3 to 4.
So the limit here is
actually that easy.
It's 3/4.
So what you do is you just
figure out what's the
fastest-growing term on the
top, what's the fastest-growing
term on the bottom, and then
figure out what it approaches.
If they're the same term, then
they kind of cancel out, and
you say the limit
approaches 3/4.
It's a very nonrigorous way
of doing it, but it gets
you the right answer.
See you in the next
presentation.