### Video Transcript

Find the limit as π₯ approaches
zero of seven π to the five π₯ minus seven over negative π to the eight π₯ plus
one.

We will start by trying to solve
this limit using direct substitution. We obtain seven times π to the
power of five times zero minus seven over negative π to the power of eight times
zero plus one. Since π to the power of zero is
equal to one, we find that this is equal to seven minus seven over negative one plus
one, which simplifies to zero over zero. However, this is undefined. Although we obtain that our limit
is undefined using direct substitution, it is equal to zero over zero. And this tells us that we may be
able to use LβHopitalβs rule.

LβHopitalβs rule tells us that if
the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to zero over zero,
positive infinity over positive infinity, or negative infinity over negative
infinity. Where π is a real number, positive
infinity, or negative infinity. 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₯.

Now our limit satisfies the
condition of our limit being equal to zero over zero. And since weβre taking the limit as
π₯ approaches zero, that means that our π is equal to zero, which is a real
number. Therefore, we can use LβHopitalβs
rule. π of π₯ is the numerator of the
function which weβre taking the limit of. So thatβs seven π to the five π₯
minus seven. And π of π₯ is the denominator. So thatβs negative π to the power of
eight π₯ plus one.

Now we must find π prime of π₯ and
π prime of π₯. Since we will be differentiating
exponential terms, we can use the rule that tells us that the differential of π to
the power of ππ₯ with respect to π₯ is equal to π times π to the power of
ππ₯. Letβs differentiate π of π₯ term
by term. Seven π to the power of five π₯ is
an exponential term. So weβll be using the rule which
weβve just stated. Our value of π is five. And we notice that we have a
constant of seven multiplying our exponential term. So that must remain two, giving us
seven timesed by five π to the five π₯. Then seven times five is 35. So we can write this as 35π to the
power of five π₯.

The second term in π of π₯ is
negative seven, which is simply a constant. And when we differentiate any
constant, we simply get zero. So we found that π prime of π₯ is
equal to 35π to the power of five π₯. The first term in π of π₯ is
negative π to the power of eight π₯, which is again an exponential term. Using our rule, we obtain that the
differential of this term is negative eight π to the power of eight π₯. The second term in π of π₯ is one,
which is again a constant. And so this will differentiate to
give zero.

We are now ready to apply
LβHopitalβs rule. We obtain that the limit as π₯
approaches zero of seven π to the power of five π₯ minus seven over negative π to
the power of eight π₯ plus one. Is equal to the limit as π₯
approaches zero of 35 times π to the power of five π₯ over negative eight times π
to the power of eight π₯. And we can now apply direct
substitution, giving us 35 timesed by π to the power of zero over negative eight
timesed by π to the power of zero. Since π to the power of zero is
equal to one, we get a solution that our limit must be equal to negative 35 over
eight.