Question Video: Finding the Limit of a Difference of Powers Mathematics

Find lim_(π‘₯ β†’ βˆ’1) (((π‘₯Β² βˆ’ 5π‘₯ βˆ’ 6)/(π‘₯ + 1)) + ((π‘₯⁷ + 1)/(π‘₯Β³ + 1))).

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Video Transcript

Find the limit as π‘₯ approaches negative one of π‘₯ squared minus five π‘₯ minus six all divided by π‘₯ plus one plus π‘₯ to the seventh power plus one divided by π‘₯ cubed plus one.

In this question, we’re asked to evaluate the limit of the sum of two rational functions. And we recall we can attempt to evaluate the limit of rational functions by using direct substitution. Substituting π‘₯ is equal to negative one into this function, we get negative one squared minus five times negative one minus six all divided by negative one plus one plus negative one to the seventh power plus one all divided by negative one cubed plus one. And then attempting to evaluate these expressions, we get zero over zero plus zero over zero. These are indeterminate forms. So we can’t just use direct substitution to evaluate this limit. And in fact we can’t even just use the fact that the limit of a sum is equal to the sum of the limits. Because if we try to apply direct substitution to each term, we also get an indeterminate form.

Instead, let’s take a closer look at the two terms inside of our limit. Let’s start with the first term. The first term is a rational function. It’s a quadratic divided by a linear function. And we know when we substitute π‘₯ is equal to negative one into our numerator, we get zero. So by using the remainder theorem, π‘₯ plus one must be a factor of our numerator. This would then allow us to evaluate the limit as π‘₯ approaches negative one of our first term.

And in fact we could do something very similar for our second term. However, there’s an easier method. We can just notice this is in the form of the limit of a difference of powers. This will then allow us to evaluate the limit of each term individually. So we’ll start by writing the limits of a sum of two functions as the sum of their limits. And of course this result is only valid provided these two limits exist.

We’ll start by evaluating the limit of the first function. We’ll do this by factoring our numerator. By the remainder theorem, we know that π‘₯ plus one must be a factor of this quadratic. We can find the other factor. It’s equal to π‘₯ minus six. Using this expression, we can rewrite the limit of the first term as the limit as π‘₯ approaches negative one of π‘₯ plus one multiplied by π‘₯ minus six all divided by π‘₯ plus one.

And now we see we’re taking the limit as π‘₯ approaches negative one of this function. Remember, we want to see what happens to the outputs of our function as π‘₯ approaches negative one. We don’t mind what happens when π‘₯ is equal to negative one. In particular, if π‘₯ is not equal to negative one, then π‘₯ plus one divided by π‘₯ plus one is just equal to one. So canceling the shared factor of π‘₯ plus one in the numerator and denominator will not change the limit of this function as π‘₯ approaches negative one.

Therefore, the limit of this first term is just the limit as π‘₯ approaches negative one of π‘₯ minus six. And since this is a linear function, we can evaluate this limit by direct substitution. We substitute π‘₯ is equal to negative one into the linear function to get negative one minus six, which is equal to negative seven.

However, we’re still not done yet. We still need to evaluate the limit of our second term. And to evaluate this limit, we need to recall the following result for the limit of a difference of powers. This tells us, for any real constants 𝑛, π‘š, and π‘Ž, where π‘š is nonzero, the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ to the π‘šth power minus π‘Ž to the π‘šth power is equal to 𝑛 over π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š. And this is provided π‘Ž to the 𝑛th power, π‘Ž to the π‘šth power, and π‘Ž to the power of 𝑛 minus π‘š all exist.

Before we apply this result, we need to rewrite our function in this form. First, we rewrite our numerator as π‘₯ to the seventh power minus negative one all raised to the seventh power. And similarly, we rewrite our denominator as π‘₯ cubed minus negative one all cubed. This means we’ve now rewritten the limit of our second term as the limit as π‘₯ approaches negative one of π‘₯ to the seventh power minus negative one to the seventh power all divided by π‘₯ cubed minus negative one all cubed. And we can see this is in the form of our limit result. π‘Ž is equal to negative one, 𝑛 is equal to seven, and π‘š is equal to three.

And now we can evaluate the limit of this second term by substituting these values into our formula. Its limit is seven over three times negative one to the power of seven minus three. We can then evaluate this expression. Seven minus three is equal to four. So we just have seven over three multiplied by negative one to the fourth power. And negative one raised to the fourth power is just equal to one.

Therefore, we’ve shown the limit of the first term in our function is equal to negative seven, and the limit of the second term in our function is equal to seven over three. So the limits of both of these functions exist. Therefore, the limit of the sum of these functions is equal to the sum of the limits. And if we add these two values together, we get negative 14 over three, which is our final answer.

Therefore, we were able to show the limit as π‘₯ approaches negative one of π‘₯ squared minus five π‘₯ minus six all divided by π‘₯ plus one plus π‘₯ to the seventh power plus one over π‘₯ cubed plus one is equal to negative 14 over three.

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