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Video: Solving Limits by Transforming Them into the Natural Exponent Limit Forms

Alex Cutbill

Determine lim_𝑥 → ∞(1 + (7/(𝑥 + 9)))^(𝑥 + 3).

10:51

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.