### Video Transcript

Find the limit as π₯ approaches
zero of 19 to the power of π₯ minus one over the square root of π₯ plus 25 minus
five.

If we try to evaluate this limit by
finding the limit as π₯ approaches zero of 19 to the power of π₯ minus one over the
limit as π₯ approaches zero of the square root of π₯ plus 25 minus five by direct
substitution of π₯ equals zero, we would end up with one minus one over five minus
five. Because, recall, that any constant
raised to the power of zero is one. So we end up with zero over zero,
an indeterminate form, which is not very helpful to us. And so we need to evaluate this
limit using LβHΓ΄pitalβs rule.

LβHΓ΄pitalβs rule says 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₯ is positive or
negative infinity and the limit as π₯ approaches π of π of π₯ is positive or
negative infinity, then the limit as π₯ approaches π of π of π₯ over π of π₯ is
the limit as π₯ approaches π of π prime of π₯ over π prime of π₯.

Remember, we got zero over zero
when we tried to evaluate this limit. So, weβre in this first
scenario. Because 19 to the power of π₯ minus
one is our π of π₯, and the square root of π₯ plus 25 minus five is our π of π₯,
so weβre going to apply LβHΓ΄pitalβs rule. And Iβm gonna clear some space on
the screen. But Iβm just going to keep this
bit, as this is what weβre going to need.

Okay, so we said that π of π₯ is
19 to the power of π₯ minus one. And we want to find π prime of
π₯. So what we need to remember here is
that if π¦ equals π to the power of π₯, then dπ¦ by dπ₯ equals π to the power of
π₯ ln π. And so π prime of π₯ equals 19 to
the power of π₯ ln 19. Because, remember, one is a
constant, so it differentiates to zero. So, thatβs π prime of π₯. Remember π of π₯ is the square
root of π₯ plus 25 minus five. And here we have a function of a
function. So, weβre going to need the chain
rule.

Iβm just going to rewrite this
slightly because, remember, taking the square root of something is the same as
raising it to the power of a half. So, this is π₯ plus 25 to the power
of a half minus five. And, remember, that the chain rule
says that if π¦ equals π of π’ and π’ equals π of π₯, then dπ¦ by dπ₯ equals dπ¦
by dπ’ multiplied by dπ’ by dπ₯.

So, letβs go ahead and
differentiate π₯ plus 25 to the power of a half using the chain rule. If we start by letting π¦ equal π₯
plus 25 to the power of a half. So, π’ is going to be π₯ plus
25. And differentiating that, with
respect to π₯, gives us that dπ’ by dπ₯ equals one. And because we let π’ equal π₯ plus
25, then we can say that π¦ equals π’ to the power of a half. And by multiplying by the power and
then subtracting one from the power with usual differentiation rules that we know,
we find that dπ¦ by dπ’ equals a half π’ to the power of negative a half.

And by the formula for the chain
rule, we have that dπ¦ by dπ₯ equals a half π’ to the power of negative a half
multiplied by one, which is just a half π’ to the power of negative a half. And, remember, we said that π’ is
equal to π₯ plus 25. So, we can replace π’ to get a half
π₯ plus 25 to the power of negative a half. And, remember, that raising
something to the power of negative a half is the same as doing one over the square
root of it. And from here, we can just tidy
this up a little bit more by multiplying these two fractions together, to give us
one over two multiplied by the square root of π₯ plus 25. And that gives us our π prime of
π₯.

So, now that we have our π prime
of π₯ and our π prime of π₯, we can apply LβHΓ΄pitalβs rule. So, Iβll clear some space for us to
do that. So, from LβHΓ΄pitalβs rule, we have
that the limit, as π₯ approaches zero, of 19 to the power of π₯ minus one over the
square root of π₯ plus 25 minus five is equal to the limit, as π₯ approaches zero,
of 19 to the power of π₯ ln 19 over one over two multiplied by the square root of π₯
plus 25.

And we know from our rules of
dividing fractions that this is the same as the limit, as π₯ approaches zero, of 19
to the power of π₯ ln 19 multiplied by two multiplied by the square root of π₯ plus
25, all over one. Which is just the limit, as π₯
approaches zero, of 19 to the power of π₯ ln 19 multiplied by two multiplied by the
square root of π₯ plus 25. And we can evaluate this limit by
directly substituting π₯ equals zero. We remember that any constant to
the power of zero is just one. And we know that the square root of
25 is five. So, finally, we end up with ln 19
times 10, which we usually write as 10 ln 19. So, thatβs our final answer.

Finally, letβs just check that we
met the conditions for LβHΓ΄pitalβs rule. π and π are differentiable
because we managed to differentiate them. Around the limit point, π₯ is zero,
π prime of π₯ is not equal to zero. And, finally, the limit, as π₯
approaches π, of π prime of π₯ over π prime of π₯ exists because we managed to
find it. So, all conditions were
satisfied.