### Video Transcript

Use the squeeze theorem to evaluate
the limit of two π squared cos of one over π as π approaches zero.

Letβs recall the squeeze
theorem. The squeeze theorem says that if
there exist functions π of π₯, π of π₯, and β of π₯ such that π of π₯ is greater
than or equal to π of π₯ but less than or equal to β of π₯ when π₯ is near π,
except possibly at π. And the limit as π₯ approaches π
of π of π₯ equals the limit as π₯ approaches π of β of π₯ equals πΏ. Then the limit as π₯ approaches π
of π of π₯ equals πΏ.

In order to use the squeeze
theorem, we need to find a function π of π₯ that we can bound from above and below
by functions β of π₯ and π of π₯, respectively. The limit we are asked to evaluate
in the question contains the function cos of one over π. Note that the value cos of π₯ is
always less than or equal to one and greater than or equal to negative one for any
of real number π₯. And so the function cos of one over
π is bounded from above by the constant function one and bounded from below by the
constant function negative one.

Now letβs multiply these
inequalities by two π squared. Doing so, we obtain two π squared
cos of one over π, which is the function whose limit weβre asked to evaluate as π
approaches zero, is less than or equal to two π squared and greater than or equal
to negative two π squared. Note that these bounds hold for any
real number π near zero apart from π equals zero, as one over π is not defined
when π is equal to zero.

As a result, letting the function
π equal to π squared cos of one over π, the function π equal negative two π
squared, the function β equal two π squared, and π equal zero in the squeeze
theorem. We find that if the limit of
negative two π squared as π approaches zero equals the limit of two π squared as
π approaches zero equals πΏ. Then the limit of two π squared
cos of one over π as π approaches zero also equals πΏ.

Letβs see if we can evaluate the
limit as π approaches zero of negative two π squared and the limit as π
approaches zero of two π squared. Firstly, note that we can factor
the multiplicative constant out of a limit. Therefore, the limit of negative
two π squared as π approaches zero is equal to negative two multiplied by the
limit of π squared as π approaches zero. For the same reason, the limit of
two π squared as π approaches zero is equal to two multiplied by the limit of π
squared as π approaches zero.

Now letβs evaluate the limit as π
approaches zero of π squared. In order to do this, we will use
the fact that the limit of π₯ to the power of π as π₯ approaches π is equal to π
to the power of π for all positive integers π. Substituting π equals zero and π
equals two into this result, we find that the limit of π squared as π approaches
zero is equal to zero squared, which is just zero.

We therefore find that the limit of
negative two π squared as π approaches zero is equal to negative two multiplied by
zero, which equals zero. And the limit of two π squared as
π approaches zero is equal to two multiplied by zero, which also equals zero. So the limit of negative two π
squared as π approaches zero is equal to the limit of two π squared as π
approaches zero. And they are both equal to
zero.

Therefore, by the squeeze theorem,
the limit of two π squared cos of one over π as π approaches zero is also equal
to zero, which is our final answer.