I've been asked to do some more limit problems.
So let's do some more.
So let's see.
What's the first problem I have?
It says the limit as x approaches 0 of x minus 2 times
the absolute value of x over the absolute value of x.
Now what immediately confuses most people is this
absolute value thing.
Because how do you deal with it?
Because you can't just subtract.
If this was an x, it would be a simple problem.
Because then you could just simplify it.
You could subtract this 2x from this x and then divide by x.
And you'd get an answer.
But this has an absolute value, so you can't just
directly manipulate it.
Well what if we're able to get rid of this absolute value?
How do we do that?
Well, in order for this limit to exist-- in order for
this limit-- I don't know.
Let's say it equals l, right?
And l is what we have to solve for.
In order for that limit to exist, then the limit to x
approaches 0 also has to exist from the positive
and the negative sides.
What do I mean?
Well that means that this must also be true.
The limit as x approaches 0 from the right-hand side, from
positive numbers, of x minus 2 times the absolute value of x
over the absolute value of x has to be equal to l.
And then we also know that the limit as x approaches 0 from
the negative side of the same function, 2 times absolute
value of x over the absolute value of x, is also going
to be equal to l, right?
If this-- If we-- As we approach 0 from the left,
we get some number.
And as we approach 0 from the right, we get the same number,
then we know that no matter what direction we approach 0
from, which is this situation, we approach that number.
So let's figure out these two limits.
And if they have the same answer, then we have the answer
to our original problem.
So how can we simplify this as x approaches 0 from
the positive side?
Well that's the same thing as the limit as x approaches
0 from the positive side.
Well if we know we're approaching 0 from the positive
side, what do we know about x?
We know that x is positive, right?
It's approaching from the right.
So if x is positive, we can get rid of these
absolute value signs.
So we just get x minus 2x over x.
And so that equals the limit as x approaches 0 from
the positive side.
Let's see.
That's x minus 2x.
This is minus x, right?
x minus x.
And then minus x divided by x is minus 1.
So the limit as x approaches 0 from the positive
side of minus 1.
Well that's just going to equal minus 1.
OK.
Now let's take this case down here when we're approaching
from the negative side.
So how can we think about this?
What is the absolute value of x if x is a negative number?
Well the absolute value of x, if x is a negative number,
is going to be the negative of x, right?
If x-- Think of it this way: if x is negative 1, when we take
the absolute value of x, we're essentially just multiplying
x times negative 1 again.
So another way to rewrite this is: this is equal to the limit
as x approaches 0 from the negative side of
x minus 2 times what?
The absolute value of x is the same thing as negative x.
Hopefully that makes sense to you, right?
If we're dealing with negative numbers, then taking the
absolute value of a negative number is the same thing as
multiplying that negative number times the negative to
essentially make it into a positive number, right?
And then, of course, the absolute value of x in the
denominator, since we're dealing with a negative number,
is also equal to negative x.
And let's see if we can simplify that.
So that means that the limit as x approaches 0 from the
negative side-- Let's see.
I get x minus 2 times negative x.
So these negatives cancel out.
So I get x plus 2x, right?
So I get 3x over minus x.
That equals minus 3.
So this is interesting.
I am approaching a different number when I approach it
from the left-hand side.
When I approached this function from the left-hand side or
from the right-hand side.
So it looks like this limit doesn't exist.
Let's confirm that.
Let me actually-- Let me see if I can-- Let me get the graphing
calculator and confirm.
Let me type it in.
So x minus-- I'll show you what I'm doing --x-- so you
don't get bored --minus absolute value of x.
And what am I-- Divided by the absolute value of x.
Let's see.
Graph.
And what does it have here?
Zoom out.
So this is-- Is this right?
x minus the absolute value of x-- Oh, sorry.
It's x minus 2 times the absolute value of x.
Graph.
There you go.
Even the-- oh.
I don't think you can see it yet.
--the graphing calculator confirms the work we did.
Although it connects them and it makes you think the it
somehow approaches here.
But that's just because it picks points and
just plots them.
So as you see, as you approach from the right-hand side, you
approach negative 1, right?
Well actually you're at negative 1 the whole time
on the right-hand side.
And as you approach from the left-hand side, you
approach negative 3.
So the limit does not exist at x is equal 0.
You would say l is undefined.
You know, there's a little dirty secret about limits.
It's not a dirty secret, but in theory you should never
get a limit problem wrong.
And why?
Well you should be able to solve it analytically.
But if you don't know how to solve it analytically, just
put it really really small numbers here.
Try out-- If you put in-- Try 0.0001.
Try numbers that are just slightly larger then whatever
your limit number is.
And then slightly smaller.
And then, just numerically, if you have a calculator, see
what it's approaching.
And sometimes you don't even need a calculator.
You could probably calculate this in your head with 0.01
or something like that.
Anyway, let's do another problem.
Invert colors.
OK.
So let's do the limit as x approaches 0 of
sine of 5x over 2x.
Now this looks a lot like sine of x over x.
And we know sine of x over x.
What is that?
And if you don't believe me, watch the videos where we prove
it using the squeeze theorem.
We proved that the limit as x approaches 0 of sine of
x over x is equal to 1.
Now this looks almost like that.
It would be great if we could get it into that form
and then we'd be done.
So how can we do that?
Well let's try to substitute.
Let's try to get a single variable here instead of
a 5 times a variable.
So let's make a substitution in magenta.
And say that a is equal to 5x.
And then that also means that-- So divide both sides by 5 and
we get x is equal to a over 5.
So let's make that substitution and do this.
So this-- The limit as x approaches 0 would be
the same thing as what?
If approaches-- As x approaches 0 here, a is also going
to approach 0, right?
So this is the same thing as the limit as a approaches 0.
Or you could view it this way: as a approaches 0, x
is still approaching 0.
Maybe at 1/5 the pace, but it's the same thing.
[SIDE COMMENTS]
OK.
Anyway, this should say the limit is a approaches 0, right?
And hopefully you're satisfied that the limit as a approaches
0 is the same-- That x is still approaching 0, right?
Because if a approaches 0 here, that's going
to make x approach 0.
So it's the limit as a approaches 0 of sine
of a over 2 times x.
But x is a over 5.
So it's 2 times a over 5.
And so that is equal to the limit as a approaches 0
of sine of a over 2/5 a.
Well that's the same thing.
We could take this out.
We could take this 1 over 2/5-- right, it's in
the denominator --out.
This is just a constant term, so we can take
it out of the limit.
And obviously if we're taking it out of the denominator,
it flips, right?
Because this is just 1 over 2/5 or 5/2.
So that equals 5/2-- right, I just took this out of the
denominator and it flips, right, because it's 1 over 2/5
--times the limit as a approaches 0 of
sine of a over a.
And now that looks an awful lot like this.
It's just we have an a instead of an x, but that doesn't
make a difference.
So this is equal to 1.
So this whole thing, since this is equal to 1, is equal to 5/2.
And, once again, if you get this answer and you're not so
sure, take your calculator out and try-- calculate.
What is the sine of 5 times 0.001?
So sine of 0.005 divided by 2 times 0.001, so 0.002.
If you take that, you're going to get a number that's awfully
close to this number.
This is approximate.
This is exactly 2.5.
You're probably going to get like 2.49999 or
something like that.
So let's do another one.
OK.
So I have the limit as x approaches 0-- this one looks a
lot like the previous one, although we have some exponents
here --of sine squared of x over x squared.
And that almost looks like sine of x over x, but I
have these squared terms.
What can I do?
Well this is the same thing.
This is sine of x squared over x squared.
So this is the same thing as the limit as x approaches 0
of sine of x over x squared, right?
If you were to take this out, you'd square the numerator,
square the denominator, and you'd get this.
If you took-- If you wanted to change this.
I won't say simplify.
Well now this is interesting.
Well this 2 is-- This is a constant term.
This isn't going to change with the limit.
So essentially, we can say, well this is going to be equal
to the limit as x approaches 0 of sine of x over x, the
whole thing squared.
And how is that?
I couldn't have done this if there was an x here, right,
because that would have changed the limit.
But this is a constant term.
So this-- The base is the only thing that's going to change as
I take-- as x approaches 0.
So I can take the limit within the base, I guess you
could say, and get here.
And now we're back to what we proved in a previous video.
What is the limit as x approaches 0 of
sine of x over x?
Right.
It's 1.
So what's 1 squared?
It is 1.
There you go.
And I've taken 13 minutes of your time.
Hopefully you've found it vaguely Useful.