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

Video Transcript

Determine the limit as 𝑥 approaches ∞ of one minus seven over 𝑥 all raised to the power of five 𝑥.

In this question, we’re asked to evaluate a limit. And we might be tempted to try and doing this directly. And if we were to try and do this directly, we would get a problem. Negative seven over 𝑥 as 𝑥 is approaching ∞ is approaching zero and five 𝑥 is approaching ∞. So, this limit is approaching one to the power of ∞, which we know is an indeterminate form. So, we’re going to need to try a different method to evaluate this limit. In fact, there are two different methods to evaluate this limit. We’ll only go through one of them.

We know that the limit result given to us the question is very similar to our limits involving Euler’s number 𝑒. And in fact, we could use either of these two definitions to evaluate our limit. It’s personal preference which one you’d use. However, usually one of the two limits is easier than the other one. And it’s very difficult to see just from the limit you’re asked which one to use. So, if you get stuck using one of the definitions, try using the other one. So, we want to try and rewrite our limit in the following form. And, of course, we can immediately see some problems.

First, inside of our parentheses, instead of having 𝑛, we have negative seven over 𝑥. So, to get around this problem, we’re just going to try setting 𝑛 equal to negative seven over 𝑥. We then want to use this substitution to rewrite our limit. We’ll start by rewriting negative seven over 𝑥 as 𝑛. This gives us the limit as 𝑥 approaches ∞ of one plus 𝑛 all raised to the power of five 𝑥. Of course, we want to rewrite this limit in terms of 𝑛. So, let’s find an expression for five 𝑥. And to do this, we can just rearrange our expression 𝑛 is equal to negative seven over 𝑥. We can multiply both sides of this equation through by 𝑥 and then divide through by 𝑛. We get 𝑥 is equal to negative seven over 𝑛. And we can then substitute this into our limit.

This gives us the limit as 𝑥 approaches ∞ of one plus 𝑛 all raised to the power of five times negative seven over 𝑛. And in fact, we can simplify our exponent. It’s equal to negative 35 over 𝑛. So, we now need to evaluate the limit as 𝑥 approaches ∞ of one plus 𝑛 all raised to the power of negative 35 over 𝑛. However, this is a problem. We have 𝑥 approaching ∞. We want to know what happens to 𝑛. To find this out, we remember that we set 𝑛 equal to negative seven over 𝑥. So, we can just ask the question “what happens to 𝑛 as 𝑥 approaches ∞?” As 𝑥 is approaching ∞, the denominator of this expression is growing without bound. So, 𝑛 is a negative number with smaller and smaller magnitude; 𝑛 is approaching zero from the left. So, we can just rewrite this as the limit as 𝑛 approaches zero from the left of one plus 𝑛 all raised to the power of negative 35 over 𝑛.

And it’s worth pointing out we don’t technically need the fact that 𝑛 approaches zero from the left. We can just write this as 𝑛 approaches zero. It doesn’t actually change the value of this limit. And now, this is almost in a form we can evaluate by using our limit result. All, we need to do is write this in terms of being to the power of one over 𝑛. First, we’ll rewrite our limit by using our laws of exponents. It’s equal to the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 all raised to the power of negative 35.

Now, all we need to do is take this exponent of negative 35 outside of our limit. And to do this, we will use the power rule for limits. The limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑛 raised to the 𝑚th power is equal to the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑛 all raised to the power of 𝑚. And this is true provided 𝑚 as an integer and the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑛 exists. And in this case, we know it’s true. We know it’s equal to Euler’s constant 𝑒. Therefore, we can just take the exponent of negative 35 outside of our limit. Then, we can just evaluate this limit to give us 𝑒. So, our answer was 𝑒 to the power of negative 35. And we can rewrite this as our final answer: one over 𝑒 to the 35th power.