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

Video Transcript

Determine the limit as 𝑥 approaches ∞ of three 𝑥 minus 11 all over three 𝑥 plus 11 all raised to the power of 𝑥.

In this question, we’re asked to evaluate a limit. And the first thing we should always try when we’re asked to evaluate a limit is to do this directly. In this case, our limit is as 𝑥 is approaching ∞. So let’s start by seeing what happens inside our parentheses. Inside our parentheses, we have a rational function. And we can see as 𝑥 approaches ∞, both the numerator and denominator of this expression are going to approach ∞. So evaluating it this way is going to give us ∞ over ∞, which we know is an indeterminant form.

So instead, we’re going to need to notice that both the numerator and denominator of this rational function are linear. So we can evaluate this by taking the quotient of the leading terms. We have three 𝑥 divided by three 𝑥 is equal to one. So this tells us as 𝑥 is approaching ∞, inside our parentheses our rational function is approaching one. And this is the exact same result we would get by just dividing by both our numerator and denominator through by 𝑥 and then evaluating the limit directly.

Our exponent is much easier to evaluate since it’s just equal to 𝑥. So it’s approaching ∞ as 𝑥 approaches ∞. So by evaluating this limit directly, we get one to the power of ∞, which is an indeterminant form. This means we can’t evaluate this limit directly, which means we’re going to need to try a different method to evaluate it.

To do this we’ll start by simplifying the expression inside of our parentheses. We could do this by using algebraic division. However, there is a simpler method. If we write the numerator of our rational function as three 𝑥 plus 11 minus 22, then we can divide three 𝑥 plus 11 and negative 22 separately by our denominator. Dividing three 𝑥 plus 11 by three 𝑥 plus 11, we get one. And dividing negative 22 by three 𝑥 plus 11, we get negative 22 over three 𝑥 plus 11.

And it’s worth pointing out here we know that three 𝑥 plus 11 is not equal to zero because 𝑥 is approaching ∞. So eventually, 𝑥 will be bigger than negative 11 over three. So we were able to rewrite our limit as the limit as 𝑥 approaches ∞ of one minus 22 over three 𝑥 plus 11 all raised to the power of 𝑥. And at this point, we have a few different methods we could use to evaluate this limit. In this video, we’re going to use one of our limit results involving Euler’s constant 𝑒.

We know the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 is equal to 𝑒. And it’s worth pointing out we have two different limit results involving Euler’s constant 𝑒. And in actual fact, because these are equivalent, we can use either of them to answer this question. It’s personal preference which one you would use. However, usually, one of the two limit results will involve easier working-out.

And it’s very difficult to see which of the two limit results you should use just by looking at the limit you’re asked to evaluate. So if you’re struggling, try switching to the other limit result. So to try and apply this limit result, inside our parentheses we’re going to want to add a value of 𝑛. This means we’re going to need to try using the substitution 𝑛 is equal to negative 22 divided by three 𝑥 plus 11. But this isn’t the only thing we need to do. Our limit is as 𝑥 is approaching ∞. If we’re using the substitution for 𝑛, we want to know what happens to 𝑛 as 𝑥 approaches ∞.

And in fact, we can do this directly from our substitution. As 𝑥 is approaching ∞, the numerator of the right-hand side of this equation remains constant. However, the denominator is approaching ∞. It’s growing without bound. So the right-hand side of this equation is approaching zero. This means 𝑛 approaches zero as 𝑥 approaches ∞. Therefore, instead of taking the limit as 𝑥 approaches ∞, we can take the limit as 𝑛 approaches zero. But there is one more thing. 𝑥 still appears in our limit. So we’re going to need to write this in terms of 𝑛.

And to do this, we can just rearrange our substitution to give 𝑥 in terms of 𝑛. We’ll start by multiplying both sides of our equation through by three 𝑥 plus 11 and then divide through by 𝑛. This gives us three 𝑥 plus 11 is equal to negative 22 over 𝑛. We want to rearrange this equation for 𝑥. We’re going to need to subtract 11 from both sides of our equation and then divide through by three. And this gives us that 𝑥 is equal to one-third times negative 22 over 𝑛 minus 11. However, we’ll simplify this by distributing one-third over our parentheses, giving us 𝑥 is equal to negative 22 over three 𝑛 minus 11 over three. And we’ll substitute this directly into our limit.

Therefore, by using our substitution, we were able to rewrite our limit as the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 22 over three 𝑛 minus 11 over three. And now we can see this is very close to our limit result. Our limit has 𝑛 approaching zero. And inside our parentheses, we have one plus 𝑛. All we need to do now is write our exponent in the form one over 𝑛.

And to do this, we’re going to want to simplify the exponent in our limit. There’s a few different ways of doing this. One way will be to use our laws of exponents. 𝑎 to the power of 𝑏 minus 𝑐 is equal to 𝑎 to the power of 𝑏 multiplied by 𝑎 to the power of negative 𝑐. And it’s worth pointing out here the only reason we didn’t write this in the form 𝑎 to the power of 𝑏 divided by 𝑎 to the power of 𝑐 is to save space. You can use either of these two methods. Both of them will work.

So by setting 𝑎 equal to one plus 𝑛, 𝑏 equal to negative 22 over three 𝑛, and 𝑐 equal to 11 over three, we get the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 22 over three 𝑛 multiplied by one plus 𝑛 all raised to the power of negative 11 over 3. Now we can see we’re trying to evaluate the limit of our product. So we can try simplifying this by using the product rule for limits.

We recall the product rule for limits tells us the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑛 times 𝑔 of 𝑛 is equal to the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑛 multiplied by the limit as 𝑛 approaches 𝑎 of 𝑔 of 𝑛 provided both the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑛 and the limit as 𝑛 approaches 𝑎 of 𝑔 of 𝑛 exist. So by applying the product rule for limits to this question, we get the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 22 over three 𝑛 multiplied by the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 11 over three. And we can only guarantee this is true provided both of these two limits exist.

And at this point, we don’t know whether both of these limits exist or not. We can directly evaluate the limit of our second function by noticing something interesting. This is a continuous function, so we can just substitute 𝑛 is equal to zero into our function. So our second limit evaluates to give us one plus zero all raised to the power of negative 11 over three, which is just equal to one. So we can just set this equal to one. This gives us the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 22 over three 𝑛 provided this limit exists. Now we’re almost in a form where we can use our limit result. We just need to write the exponent in the form one over 𝑛.

And we can do this by using our laws of exponents. We’re going to use the fact that 𝑎 to the power of 𝑏 times 𝑐 is equal to 𝑎 to the power 𝑏 all raised to the power of 𝑐. So by setting 𝑎 equal to one plus 𝑛, 𝑏 equal to one over 𝑛, and 𝑐 equal to negative 22 over three, we’ve rewritten our limit as 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 22 over three provided this limit exists.

And now it might be tempting to just try directly applying our limit result. However, we can’t do this yet because our exponent is still inside of our limit. So we want to take this exponent of negative 22 over three outside of our limit. And to do this, we’re going to need to use the power rule for limits. We recall the power rule for limits tells us the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑥 raised to the power of 𝑘 is equal to the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑥 all raised to the power of 𝑘 provided the limit as 𝑛 approaches 𝑎 of 𝑓 of 𝑥 exists and raising this to the power of 𝑘 has to also exist.

And in our case, we’ll be able to prove that both of these prerequisites are true. The easiest way to do this is to apply the power rule for limits to our limit and then check both of the prerequisites are true. So 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 22 over three. We can see that our limit inside of our exponent exists. In fact, we know that this is equal to 𝑒 because this is just our limit result.

Next, we can also see that raising this to the power of negative 22 over three also exists because 𝑒 is just a positive number. This justifies our use of the power rule for limits, giving us 𝑒 to the power of negative 22 over three. And it is also worth pointing out because we’ve just shown that this limit exists, we’ve also justified our use of the product rule for limits. Therefore, we’ve successfully evaluated our limits, so we could leave our answer like this. However, we’ll do one more piece of modification. We’re going to use our laws of exponents to rewrite this in the denominator. We get one over 𝑒 to the power of 22 over three, which is our final answer.

Therefore, we were able to show the limit as 𝑥 approaches ∞ of three 𝑥 minus 11 all over three 𝑥 plus 11 all raised to the power of 𝑥 is equal to one over 𝑒 to the power of 22 over three.