### Video Transcript

Determine the limit of one plus
seven over 𝑥 plus nine all to the power of 𝑥 plus three as 𝑥 approaches ∞.

This is a tricky limit to evaluate
because the variable 𝑥, which is approaching ∞, appears in both the base and the
exponent of this power. Hopefully, you will have seen
another limit where the variable approaching ∞ appears in both the base and the
exponent, the limit of one plus one over 𝑛 all to the power of 𝑛 as 𝑛 approaches
∞. And the value of this limit is
Euler’s number, 𝑒, the base of the natural logarithm, which is approximately
2.718.

The way to find the value of our
limit is to rewrite this limit in terms of the limit whose value we know. Okay, well, how do we do that? The powers inside our limit and the
known limit have different bases. But we can change this by finding a
function 𝑛 of 𝑥 for which the base of our limit, one plus seven over 𝑥 plus nine,
is one plus one over 𝑛 of 𝑥. So let’s find the function 𝑛 of 𝑥
which has this property.

We cancel the ones or alternatively
subtract one from both sides. We multiply by 𝑛 of 𝑥 and 𝑥 plus
nine. And then we can divide by seven to
find 𝑛 as a function of 𝑥. Alternatively, we can think of 𝑥
as a function of 𝑛 by subtracting nine from both sides and then swapping the
sides.

We can substitute this expression
for 𝑥 in terms of 𝑛 of 𝑥 into the limit we want to find to get to the limit of
one plus seven over seven 𝑛 of 𝑥 minus nine plus nine all to the power of seven 𝑛
of 𝑥 minus nine plus three. We can simplify the exponent. Minus nine plus three becomes minus
six. But of course, we can also simplify
the base. We have a minus nine and a plus
nine, which cancel. And then we’re left with a fraction
of seven over seven 𝑛 of 𝑥, which is of course just one over 𝑛 of 𝑥.

Let’s tidy this up then. We’re left with the limit of one
plus one over 𝑛 of 𝑥 all to the power of seven 𝑛 of 𝑥 minus six as 𝑥 approaches
∞. And maybe we should remind
ourselves what 𝑛 of 𝑥 is. 𝑛 of 𝑥 is 𝑥 plus nine over
seven.

Now we can see that the base of the
power in our limit is the same as the base of the power in the known limit. Of course, this is no surprise
because we chose 𝑛 of 𝑥 carefully to make this happen. It’s just the exponents that are
different.

Well, actually, that’s not quite
true. Our limit is the limit as 𝑥
approaches ∞, whereas the known limit has 𝑛 approaching ∞. Our limit is the limit as 𝑥
approaches ∞ of some function, but that function only depends on 𝑥 via the function
𝑛 of 𝑥.

Let’s stop thinking about 𝑛 as a
function of 𝑥 and consider it as a variable in its own right. What does this mean for our
limit? As 𝑥 approaches ∞, what’s
happening to 𝑛? Well, 𝑛 is 𝑥 plus nine over
seven. So as 𝑥 goes to ∞, 𝑛 is also
going to ∞. And so we can replace the limit as
𝑥 goes to ∞ by the limit as 𝑛 goes to ∞. They are equivalent.

It might be worth pausing the video
here and thinking about why this is true intuitively, perhaps making use of the
expressions we found for 𝑛 in terms of 𝑥 and 𝑥 in terms of 𝑛. We can continue to ignore the fact
that 𝑛 is actually a function of 𝑥 and just think of it as some variable which has
some value.

So we just write 𝑛 where before we
wrote 𝑛 of 𝑥. And we don’t need to explain how 𝑛
is related to 𝑥. All that we need for the purposes
of evaluating this limit is to say that 𝑛 is going to ∞. We will justify this step formally
at the end of the video, using a property of the limit of a composition of
functions. But it’s still worth checking now
that this step makes intuitive sense.

Okay, so if you remember, our
strategy for finding the value of our limit was to relate it to the known limit, the
limit of one plus one over 𝑛 all to the power of 𝑛 as 𝑛 approaches ∞, whose value
we know to be Euler’s number, 𝑒. And I think we’re doing pretty well
so far. Our limit looks quite a lot like
the known limit. The only difference now is in the
exponent.

And so we have the easier task of
writing the limit of something to the power of seven 𝑛 minus six as 𝑛 approaches ∞
in terms of the limit of that something to the power of 𝑛 as 𝑛 approaches ∞. We can apply a law of exponents to
the function inside the limit. And so our limit becomes the limit
of one plus one over 𝑛 to the power of seven 𝑛 all over one plus one over 𝑛 to
the power of six as 𝑛 approaches ∞.

And we can also use the fact that
the limit of a quotient of functions is the quotient of their limits. So we have the limit of one plus
one over 𝑛 to the power of seven 𝑛 as 𝑛 approaches ∞ divided by the limit of one
plus one over 𝑛 to the power of six as 𝑛 approaches ∞. And notice that, in the limit in
the denominator, the exponent six does not depend on 𝑛. And so this limit is relatively
straightforward to evaluate.

We use the fact that the limit of a
power of a function is that power of the limit of the function, to take the exponent
outside the limit. And as the limit of one plus one
over 𝑛 as 𝑛 approaches ∞ is just one, the limit in the denominator is one to the
power of six, which of course is just one. And so we only have the limit in
the numerator to worry about.

And it may not to be immediately
obvious how we can write the limit over one plus one over 𝑛 to the power of seven
𝑛 as 𝑛 approaches ∞ in terms of the limit of one plus one over 𝑛 to the power of
just 𝑛 as 𝑛 approaches ∞. But it might be easier to see if we
swap the order of the seven and the 𝑛 in the exponent, leaving us with an exponent
of 𝑛 times seven.

We can use another law of
exponents, which says that 𝑎 to the power of 𝑏 times 𝑐 is 𝑎 to the power of 𝑏
all to the power of 𝑐. And so we have to find the limit of
one plus one over 𝑛 to the power of 𝑛 all to the power of seven as 𝑛 approaches
∞. And as we saw before, we can swap
the order of taking the limit and exponentiating, to get the limit of one plus one
over 𝑛 to the power of 𝑛 as 𝑛 approaches ∞, all to the power of seven.

And now we have succeeded in
writing our limit in terms of the limit of one plus one over 𝑛 to the power of 𝑛
as 𝑛 approaches ∞, whose value we know to be 𝑒. And so the value of our limit is 𝑒
to the power of seven.

Let’s just recap what we did. We noticed that the function we had
to find the limit of had the variable 𝑥 in both the base and exponent. And this reminded us of a limit
whose value we know, the limit of one plus one over 𝑛 all to the power of 𝑛 as 𝑛
approaches ∞, whose value is Euler’s number 𝑒.

We found a function 𝑛 of 𝑥 which
allowed us to write the base of our function as one plus one over 𝑛 of 𝑥. And we inverted the function 𝑛 to
find 𝑥 in terms of 𝑛 of 𝑥, which allowed us to replace all instances of 𝑥 in our
function by expressions involving 𝑛 of 𝑥.

Having done this, we argued that we
could forget about 𝑥, replacing 𝑛 of 𝑥 by just 𝑛 and the limit as 𝑥 approaches
∞ by the limit as 𝑛 approaches ∞. After that, we just had to apply
some laws of exponents and laws of limits to get our limit in terms of the limit of
one plus one over 𝑛 to the power of 𝑛 as 𝑛 approaches ∞. And as we know the value of this
limit is 𝑒, we just had to substitute 𝑒 to find our answer of 𝑒 to the power of
seven.

I said earlier that I would explain
more formally how the third line of working follows from the second. In the second line, we have a
composition of functions. We have the limit of 𝑓 of 𝑛 of 𝑥
as 𝑥 approaches ∞, where 𝑓 is a function which takes 𝑥 to one plus one over 𝑥
all to the power of seven 𝑥 minus six.

Now for any two functions 𝑓 and
𝑔, the limit of 𝑓 of 𝑔 of 𝑥 as 𝑥 approaches 𝑎 is the limit of 𝑓 of 𝑛 as 𝑛
approaches the limit of 𝑔 of 𝑥 as 𝑥 approaches 𝑎. Perhaps it would be easier if we
replaced 𝑛 of 𝑥 by its definition, 𝑥 plus nine over seven. So let’s do that.

Now we can apply our rule, with 𝑎
equal to ∞ and 𝑔 of 𝑥 being 𝑥 plus nine over seven. We want to evaluate the right-hand
side, but first we need to know what 𝑛 is approaching. It’s approaching the limit of 𝑥
plus nine over seven as 𝑥 approaches ∞. And as this limit is ∞, we’re left
with the limit of 𝑓 of 𝑛 as 𝑛 approaches ∞. And as 𝑓 of 𝑥 is one plus one
over 𝑥 to the power of seven 𝑥 minus six, we get the limit of one plus one over 𝑛
all to the power of seven 𝑛 minus six as 𝑛 approaches ∞. And this is the third line of
working as claimed. This more formal justification of
the third line of working used the limit law for a composition of functions.