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

Determine lim_(π‘₯ β†’ ∞) ((7π‘₯ + 5)/(7π‘₯ + 2))^(βˆ’6π‘₯ βˆ’ 5).

08:41

Video Transcript

Determine the limit as π‘₯ approaches ∞ of seven π‘₯ plus five all over seven π‘₯ plus two all raised to the power of negative six π‘₯ minus five.

In this question, we’re asked to evaluate a limit. And the first thing we should always do when we’re asked to evaluate a limit is try and do this directly. In our limit, we can see we have π‘₯ approaching ∞. And inside our parentheses, we have the quotient of two polynomials. Of course, both the numerator and denominator of this expression are approaching ∞ as π‘₯ approaches ∞. So this won’t help us. Instead, we’re going to need to notice that both the numerator and denominator are linear functions. So in particular, the degree of the polynomials are equal. And then we look at the quotients of the leading terms, in this case seven over seven, which is equal to one. So inside our parentheses, as π‘₯ approaches ∞, this is equal to one. In other words, the numerator and denominator are growing at the same rate. And the exponent of this expression is far easier to evaluate. We have negative six π‘₯ minus five. As π‘₯ approaches ∞, negative six π‘₯ is going to be approaching negative ∞.

So by trying to evaluate this limit directly, we get one to the power of negative ∞, which is an indeterminate form. So we’re going to need to try and use some other method to evaluate this limit. To evaluate this limit, we’re going to start by simplifying the expression inside our parentheses. There’s a few different ways of doing this. For example, we could use polynomial division. However, the easiest way is to rewrite the numerator of this expression as seven π‘₯ plus two plus three. Then we can see dividing seven π‘₯ plus two by our denominator of seven π‘₯ plus two gives us one. And then we get three over seven π‘₯ plus two. So we can replace the rational function in our parentheses with one plus three over seven π‘₯ plus two. This gives us the limit as π‘₯ approaches ∞ of one plus three over seven π‘₯ plus two all raised to the power of negative six π‘₯ minus five.

Now there’s a few different ways of evaluating this limit. The easiest way will be to use one of our limit results involving Euler’s constant 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, it is worth pointing out we can use the other limit result we have involving Euler’s constant 𝑒 to answer this question. But usually one of the two limit results is easier than the other one. And it’s very difficult to tell which one to use just by looking at the limit we’re asked to evaluate. So if we’re struggling, we should try switching to the other limit result. So to use this limit result to evaluate our limit, inside our parentheses, we’re going to want one plus 𝑛.

So we’re going to need to evaluate this limit by substitution. We’re going to use the substitution 𝑛 is equal to three divided by seven π‘₯ plus two. But this isn’t the only time π‘₯ appears in this limit. Our limit is as π‘₯ is approaching ∞ and π‘₯ appears in our exponent. So we’re going to want to rewrite all of our instances of π‘₯ to be in terms of 𝑛. Let’s start with π‘₯ approaching ∞. We can see as π‘₯ approaches ∞, on the right-hand side of this equation, we have three divided by seven π‘₯ plus two. Our numerator remains constant; however, our denominator is growing without bound. So as π‘₯ is approaching ∞, we must have that 𝑛 is approaching zero. In fact, we also know that 𝑛 is approaching zero from the right. However, it’s not necessary to use this piece of information.

Next, we’re going to want to find an expression for π‘₯ in terms of 𝑛. And to do this, we’re going to need to rearrange our equation for 𝑛. We’ll start by multiplying both sides of our equation through by seven π‘₯ plus two and then dividing through by 𝑛. This gives us seven π‘₯ plus two is equal to three divided by 𝑛. Remember, we want an expression for π‘₯. So we’re going to need to subtract two from both sides of this equation and then divide through by seven. This gives us π‘₯ is equal to one-seventh multiplied by three over 𝑛 minus two. We’re going to substitute this into our limit, so we can now use this to rewrite our limit. First, remember, as π‘₯ approaches ∞, 𝑛 is approaching zero. So we get the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative six multiplied by one over seven times three over 𝑛 minus two subtract five.

Remember, we’re trying to write this limit in the form of our limit result. We now have 𝑛 approaching zero in our limit, and we have one plus 𝑛 inside of the parentheses. But we need the exponent to just be in the form one over 𝑛. So all we’re going to do is simplify our exponent. And to do this, all we’re going to do is distribute one-seventh over our parentheses and distribute negative six over our parentheses. Our exponent will end up being negative 18 over seven 𝑛 plus 12 over seven minus five. And although it’s not necessary, we can simplify this expression. We have 12 over seven minus five is equal to negative 23 over seven.

So we’ve now rewritten our limit as the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 18 over seven 𝑛 minus 23 over seven. And we’re much closer to being able to directly apply our limit result. Remember, we want our exponent to just be one over 𝑛. So we want to write our limit with the exponent just being one over 𝑛. So to do this, we’ll start by splitting our limit by using our laws of exponents. There’s a few different ways of doing this. We’re going to use the law of exponents, which tells us 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. And as we’ll see, we could also do this by writing this as π‘Ž to the power 𝑏 divided by π‘Ž to the power of 𝑐.

We just choose to use this result because it will save space. So this gives us the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 18 over seven 𝑛 multiplied by one plus 𝑛 all raised to the power of negative 23 over seven. Now we can see we’re taking the limits of a product. So we’ll want to simplify this by using the product rule for limits. We recall the product rule for limits tells us the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 times 𝑔 of 𝑛 is equal to the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 multiplied by the limit as 𝑛 approaches π‘Ž of 𝑔 of 𝑛 provided both the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 and the limit as 𝑛 approaches π‘Ž of 𝑔 of 𝑛 exist.

In our case, we don’t quite know that both of these limits exist here. We do know the limit of our second factor exists at this point. As 𝑛 approaches zero, inside our parentheses, 𝑛 is going to approach zero. So this is going to approach one to the power of negative 23 over seven, which is just equal to one. Then, if the limit of our first factor does exist, by using the product rule for limits, we can rewrite our limit as the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 18 over seven 𝑛 multiplied by one. And of course, multiplying by one isn’t going to change the value of our limit. So to justify our use of the product rule for limits, we’re going to need to check that this limit exists. But we can check this by carrying on our working out and then going back if it turns out this limit does not exist.

Now we can almost use our limit result. We’re just going to need to write our exponent as one over 𝑛. But before we do this, let’s clear some space for the rest of our working. We want to rewrite this expression so the exponent is one over 𝑛. To do this, we’re going to need to use one of our laws of exponents. π‘Ž to the power of 𝑏 times 𝑐 is equal to π‘Ž to the power of 𝑏 all raised to the power of 𝑐. Applying this with π‘Ž to be one plus 𝑛, 𝑏 to be one over 𝑛, and 𝑐 to be negative 18 over seven, 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 18 over seven.

And at this point, it would be very tempting to just directly apply our limit result. However, we can’t do this yet because our exponent of negative 18 over seven is still inside of our limit. So our limits don’t perfectly match up. Instead, 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 π‘Ž of 𝑓 of 𝑛 raised to the π‘˜th power is equal to the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 all raised to the π‘˜th power, provided the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 exists. And when we raise this to the π‘˜th power, this also needs to exist. And in our case, we’ll be able to show that both of these things are true.

And the easiest way to show both of these prerequisites will be to show the next line of working. If we were to apply the power rule for limits, we would now have the limit as 𝑛 approaches zero of one plus 𝑛 raised to the power of one over 𝑛 all raised to the power of negative 18 over seven. And to justify that this is true, we need to show two things. First, we need to show that the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 exists. And this is just our limit result. We already know this is equal to Euler’s constant 𝑒. But this isn’t all. We also need that we’re allowed to raise this to the power of negative 18 over seven, which we know we can do because 𝑒 is positive.

So because this is true, we’ve justified our use of the power rule for limits. This means this limit has to be equal to 𝑒 to the power of negative 18 over seven. And it’s also worth pointing out here because we’ve shown this limit exists, we’ve also just justified our use of the product rule for limits we used earlier. Because remember, we hadn’t yet showed the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative 18 over seven 𝑛 existed. But we just showed this exists and it’s equal to 𝑒 to the power of negative 18 over seven. And we could leave our answer like this. However, we’ll use our laws of exponents to rewrite this as one over 𝑒 to the power of 18 over seven, which is our final answer.

Therefore, we were able to show the limit as π‘₯ approaches ∞ of seven π‘₯ plus five all over seven π‘₯ plus two all raised to the power of negative six π‘₯ minus five is equal to one divided by 𝑒 to the power of 18 over seven.

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