In this video, I will introduce the squeeze theorem.
I will explain what the theorem says and how to use it.
But before I do that, I want to use an exercise as motivation.
This exercise looks simple, but is deceiving; it is a trap!
I want to compute lim_(x → 0) (x^2 ⋅ sin(1/x)).
There is a quick, tempting way to answer this, and it is as follows:
this function is the product of two functions,
so the limit will be the product:
the limit of x^2 times the limit of sin(1/x).
And the first limit, x^2, is 0.
And 0 times anything is 0, so the answer is 0.
That's it.
This argument is completely WRONG.
This is an incorrect solution. And here is why:
implicitly, in this calculation, I have used the limit law for products.
Remember, the limit law for products says
that IF the limit of f is L and the limit of g is M,
THEN the limit of the product is L ⋅ M.
But in order to use this theorem,
I need to ASSUME that the two initial limits exist
-- and "exist" means "are real numbers."
And I cannot do that here.
Because in this example, sin(1/x) does NOT have a limit at 0.
When x approaches 0,
the function oscillates wildly between -1 and 1.
Therefore, I cannot use the limit law for products
and the argument is completely wrong.
In case you are not convinced,
here are a few other examples where if I use the same argument,
I will get an answer that is clearly false.
All of these limits are as x approaches 0.
First: the limit of x is 0. 3/x does NOT have a limit.
But their product has a limit, and that limit is 3.
On the other hand, second example:
the limit of x is 0, 3/(x^2) does NOT have a limit,
but their product, well, ALSO does not have a limit.
So, in the original problem:
the limit of x^2 is 0, sin(1/x) does NOT have a limit,
so their product...
I simply cannot conclude anything;
with the information I have, I don't know what the answer is yet.
I need to do something else.
Luckily, there is something I can do.
The part that is giving me trouble is sin(1/x).
And, even though it doesn't have a limit at 0,
it is still a little bit under control,
because sine of anything is ALWAYS between -1 and 1.
So when I multiply it by x^2,
the function I care about is always between -x^2 and x^2.
And this is useful!
Let's look at the graph.
Here, I have sketched the graph of the two parabolas:
y = x^2, and y = -x^2.
And the function I care about is between them.
So its graph will be squeezed or sandwiched between them,
like this picture shows.
And this picture suggests that therefore, no matter how much it oscillates
-- and it DOES oscillate --
the limit at 0 WILL exist and be 0.
This is what the squeeze theorem says.
Here is a more general picture of the same phenomenon.
If I have two functions, f and h, and they both have the same limit L at a,
and then I have a THIRD function
that is sandwiched or SQUEEZED IN BETWEEN THEM,
then no matter how it behaves,
it must also have the same limit at that point.
Let's write this out as a theorem.
IF I have three functions in order h is smaller than g is smaller than f,
AND IF I ASSUME that the small function AND the large function,
they both have the SAME limit,
THEN I can conclude
that the function in the middle also has the same limit.
To be precise, the string of inequalities
DOESN'T need to be true for ALL values of x.
I only need it to be true when x is close to a but not a.
I don't care what happens AT a.
And to make sure this is a well written theorem,
in the statement I need to begin by introducing my variables.
So before I say this if-then,
I say that a and L are real numbers,
and I require f, g, and h to be functions defined near a,
except possibly at a.
And that's it!
This is the statement of the squeeze theorem.
And the picture you should always carry in your mind to accompany it
that really well summarizes what's going on
is this one.
Now, let's use this shiny new theorem to solve the original problem.
Remember: I wanted to compute lim_(x → 0) (x^2 ⋅ sin(1/x)).
Well, with the squeeze theorem, it's actually pretty fast.
All I need to do is notice that whenever x is not 0,
this function is sandwiched between -x^2 and x^2.
AND those two new functions have limits 0 and 0.
Therefore, by the squeeze theorem, the original function ALSO has limit 0.
That's it.
So the answer was 0 all along,
but the initial argument was wrong.
This, on the other hand, is a correct argument.
There is only one thing left to do: prove the squeeze theorem.
I will do that in the next video.