### Video Transcript

For the functions π of π₯ equals
two π₯ squared and π of π₯ equals four π to the two π₯ power, evaluate the limit
as π₯ approaches β of π of π₯ over π of π₯ using LβHΓ΄pitalβs rule.

Letβs firstly remind ourselves of
LβHΓ΄pitalβs rule. Suppose that π and π are
differentiable and π prime of π₯ is not equal to zero near π, but π prime of π
equals zero is okay. If the limit as π₯ approaches π of
π of π₯ is zero and the limit as π₯ approaches π of π of π₯ is zero or the limit
as π₯ approaches π of π of π₯ is positive or negative β and the limit as π₯
approaches π of π of π₯ is positive or negative β. Then the limit as π₯ approaches π
of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π prime of
π₯ over π prime of π₯, as long as this limit exists or is positive or negative β.

Although this looks quite
complicated, all itβs saying is that as long as we meet these conditions, if we
evaluate a limit by direct substitution and get an indeterminate form, then the
limit of the quotient of these functions is the same as the limit as the quotient of
the derivatives of these functions. Now, for our question, weβve
already been told what π of π₯ and π of π₯ is equal to. Before we use LβHΓ΄pitalβs rule, we
must check that we meet the required conditions.

Firstly, we must have that π and
π are both differentiable. Well, we can differentiate π of π₯
equals two π₯ squared using the power rule. That gives us that π prime of π₯
equals four π₯. And we can differentiate π of π₯
equals four π to the two π₯ power using differentiation rules for exponential
functions. The chain rule tells us that the
derivative with respect to π₯ of π raised to a function of π₯ power is equal to π
prime of π₯ multiplied by π raised to the power of π of π₯. So π prime of π₯ is equal to eight
π to the two π₯ power. So because we managed to
differentiate π and g, weβve satisfied that first condition.

We also need to make sure that π
prime of π₯ is not equal to zero near π. π is the limit, which we weβre
given in the question as β. So if we evaluate π prime of π₯ at
β, we can see that this is not going to be zero. So weβve satisfied that condition
too. Now, letβs double-check that we get
an indeterminate form when we try to evaluate this limit directly.

The limit as π₯ approaches β of two
π₯ squared is just going to be β and the limit as π₯ approaches β of four π to the
two π₯ power is also going to be β. So we find ourselves in this
situation. What LβHΓ΄pitalβs rule tells us is
that the limit as π₯ approaches β of two π₯ squared over four π to the two π₯ power
is the same as the limit as π₯ approaches β of four π₯ over eight π to the two π₯
power. But this limit just gives us β over
β, which is another indeterminate form. So we apply LβHΓ΄pitalβs rule again,
but this time to this function. So by LβHΓ΄pitalβs rule, the limit
as π₯ approaches β of four π₯ over eight π to the two π₯ power is the same as the
limit as π₯ approaches β of four over 16π to the two π₯ power. This is because four π₯
differentiated to give us four and eight π to the two π₯ power differentiated to
give us 16π to the two π₯ power.

We know that weβre okay to use
LβHΓ΄pitalβs rule because both the function on the numerator and denominator are
differentiable and 16π to the two π₯ power is not equal to zero near β. And weβre in this scenario where
the limit as π₯ approaches β of the function on the numerator and denominator both
were β. So now, applying the limit as π₯
approaches β, we find that the denominator approaches β. So the limit of the function is
zero. And that gives us our final
answer.

So by applying LβHΓ΄pitalβs rule
twice to our function, we were able to evaluate this limit. Remember, when solving this type of
question, itβs important to check that you meet the necessary conditions to use
LβHΓ΄pitalβs rule.