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

Determine lim_(π‘₯β†’βˆž) (1 βˆ’ (6/π‘₯))^(π‘₯ βˆ’ 9).

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Video Transcript

Determine the limit as π‘₯ approaches ∞ of one minus six over π‘₯ all raised to the power of π‘₯ minus nine.

In this question, we’re asked to evaluate a limit, and the first thing we should always try and do is try and evaluate this limit directly. Our limit is as π‘₯ is approaching ∞, and we can see inside of our parentheses as π‘₯ approaches ∞, negative six over π‘₯ is approaching zero because its numerator remains constant. However, its denominator is growing without bounds. Therefore, as π‘₯ approaches ∞ inside of our parentheses, we have one minus zero which is equal to one. But remember, we still have our exponent π‘₯ minus nine. As π‘₯ approaches ∞, this is approaching ∞. So when we’re try and evaluate this limit directly, we get one to the power of ∞, which is an indeterminate form. Therefore, we can’t use this to evaluate our limit. So, we’re going to need to try and do something else to help us evaluate this limit.

To do this, we can notice the limit given to us in this question is very similar to limits which we already know how to evaluate. It’s very similar to the limit we have for Euler’s number 𝑒. In fact, to evaluate this limit, we can use either of the two limits we know evaluate to give us Euler’s number 𝑒. In this video, we’re going to use the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 is equal to 𝑒.

However, remember, whenever we’re answering questions like this, usually one of our two limit results for Euler’s number 𝑒 makes this question a lot easier. But it’s difficult to know which one just by looking at the limit we need to evaluate. So if you ever get stuck trying to use one limit, try using the other limit. To try and use this result, we’re going to want to try and rewrite our limit in this form. And to do this, we’re going to want one plus 𝑛 inside of our parentheses. So, the first thing we’re going to need to do is substitute 𝑛 is equal to negative six over π‘₯. By using this substitution, we can rewrite this limit as the limit as π‘₯ approaches ∞ of one plus 𝑛 all raised to the power of π‘₯ minus nine.

But at this point, we can’t directly evaluate this limit either. To do this, we’re going to want to rewrite our limit to be entirely in terms of 𝑛. To do this, let’s start by finding an expression for π‘₯ in terms of 𝑛. To do this, we’re going to need to rearrange our substitution. We multiply both sides of this equation through by π‘₯ and then divide through by 𝑛. This gives us that π‘₯ is equal to negative six divided by 𝑛. And now we can substitute this into our limit instead of π‘₯. This gives us the limit as π‘₯ approaches ∞ of one plus 𝑛 all raised to the power of negative six over 𝑛 minus nine.

The last thing we’re going to want to do is see what happens to our value of 𝑛 as π‘₯ approaches ∞. And to do this, the easiest way is just to look at our substitution. As our values of π‘₯ approach ∞, negative six divided by π‘₯ is going to be approaching zero. So 𝑛 is just approaching zero, so we can rewrite our limit to have 𝑛 approaching zero. This gives us the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative six over 𝑛 minus nine. And now we can see this limit is almost in the form of our limit result. The only difference here is our exponent. Instead of having the exponent of one over 𝑛, we have negative six over 𝑛 minus nine. So we’re going to want to try and rewrite this limit to have the exponent of one over 𝑛.

And to do this, we’re going to need to use our laws of exponents. First, we can see that our exponent is the difference between two values. So we’re going to use the following exponent law: π‘Ž to the power of 𝑏 minus 𝑐 is equal to π‘Ž to the power of 𝑏 multiplied by π‘Ž to the power of negative 𝑐. By using this, we can rewrite our limit as the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative six over 𝑛 multiplied by one plus 𝑛 all raised to the power of negative nine. And now we can see we’re taking the limit of a product. And we know the product of a limit is equal to the limit of the product provided the limit of both products exists. And we’ll see that this is in fact true. Both of these limits exist.

Let’s start with the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative nine. We can evaluate this by direct substitution because it’s a continuous function when 𝑛 is equal to zero. We get one plus zero all raised to the power of negative nine, which is of course just equal to one. Therefore, we get that this limit is equal to the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative six over 𝑛, provided this limit exists. And now this limit is almost in a form we can evaluate by using our limit law. We just need to write our exponent in the form one over 𝑛.

And once again, we’re going to do this by using one of our laws for exponents. This time, we’re going to use the fact that π‘Ž to the power 𝑏 over 𝑛 is equal to π‘Ž to the power of one over 𝑛 all raised to the power of 𝑏. In this case, our value of 𝑏 is going to be equal to negative six. Therefore, by using this, we get 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 six. Now we can see we’re taking the limit of a function raised to an integer power, so we can try evaluating this limit by using the power rule for limits.

We recall one version of the power rule for limits tells us the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ raised to the π‘šth power is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ all raised to the π‘šth power. And we can guarantee this is true provided π‘š is an integer and the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ exists. And in this case, both of these are true. Our value of π‘š is negative six, which is an integer. And to check that our limit exists, we need to know the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛. And in fact, this is just our limit result. We already know this is equal to 𝑒. Therefore, we can simplify this limit result by using the power rule for limits.

Instead of raising just the function inside of our limit to the power of negative six, we can instead raise our entire limit to the power of negative six. And we already know how to evaluate the limit inside of our parentheses. It’s just equal to 𝑒. So this is all just equal to 𝑒 to the power of negative six. I will just rewrite this by using our laws of exponents as one over 𝑒 to the sixth power which is our final answer. Therefore, we were able to show the limit as π‘₯ approaches ∞ of one minus six over π‘₯ all raised to the power of π‘₯ minus nine is equal to one over 𝑒 to the sixth power.

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