### Video Transcript

Find the limit as π₯ approaches two
of nine to the power of π₯ minus 81 over seven to the power of π₯ minus 49.

If we try to evaluate this limit by
evaluating the limit as π₯ approaches two of nine to the power of π₯ minus 81 over
the limit as π₯ approaches two of seven to the power of π₯ minus 49, we would get
nine squared minus 81 over seven squared minus 49. And this is 81 minus 81 over 49
minus 49. And this is just zero over zero,
which is an indeterminate form. So, instead, we evaluate this limit
by using lβHopitalβs rule.

LβHopitalβs rule states that if the
limit as π₯ approaches π of π of π₯ equals zero and the limit as π₯ approaches π
of π of π₯ equals zero. Or the limit as π₯ approaches π of
π of π₯ equals 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₯.

Remember, when we initially tried
to evaluate our limit, we got zero over zero by direct substitution. So we have this first scenario. So letβs apply lβHopitalβs rule to
our question. We can see that, for our problem,
π of π₯ is nine to the power of π₯ minus 81. And π of π₯ is seven to the power
of π₯ minus 49. So letβs go ahead and differentiate
the top function.

So whatβs the derivative of nine to
the power of π₯ minus 81? Well, one helpful rule that weβre
going to need here is that if π¦ equals π to the power of π₯, so a constant raised
to the power of π₯, then dπ¦ by dπ₯ equals π to the power of π₯ multiplied by ln
π₯. So nine to the power of π₯ minus 81
will differentiate to nine to the power of π₯ multiplied by ln nine. As 81 is a constant, it
differentiates to zero. And for the same reasons, seven to
the power of π₯ minus 49 differentiates to seven to the power of π₯ multiplied by ln
seven.

And so, by lβHopitalβs rule, the
limit as π₯ approaches two of nine to the power of π₯ minus 81 over seven to the
power of π₯ minus 49 is equal to the limit as π₯ approaches two of nine to the power
of π₯ multiplied by ln nine over seven to the power of π₯ multiplied by ln
seven. And if we now evaluate this limit
by direct substitution of π₯ equals two, we get nine squared multiplied by ln nine
over seven squared multiplied by ln seven, which is just 81 multiplied by ln nine
over 49 multiplied by ln seven.

Now, itβs important to know some
conditions for lβHopitalβs rule. Firstly π and π must both be
differentiable, which they are, as we differentiated nine to the power of π₯ minus
81 to get nine to the power of π₯ multiplied by ln nine. And we differentiated seven to the
power of π₯ minus 49 to get seven to the power of π₯ multiplied by ln seven. So that condition is satisfied.

Secondly, we must have that π
prime of π₯ is not equal to zero near π. And for our question, π prime of
π₯ was seven to the power of π₯ multiplied by ln seven, which is not zero near the
limit 0.2. So that condition is satisfied.

And finally the limit as π₯
approaches π of π prime of π₯ over π prime of π₯ must exist or be positive or
negative β. And for our question, this does
exist, because weβre able to find it. So the final condition was
satisfied.