Video Transcript
Determine the limit as 𝑥 approaches negative five of one divided by 𝑥 plus seven minus one-seventh all divided by 𝑥.
We’re asked to evaluate the limit of a very complicated-looking function. And the first thing we should always do when we’re asked to evaluate limits is check to see if we can do this by using direct substitution. In this case, we can see the function inside of our limit is not a standard function, so we’re going to need to do a little bit more work. Instead, we’re going to need to look at each part of our function separately. We’ll start with the first term in our numerator, one divided by 𝑥 plus seven.
We know that one divided by 𝑥 plus seven is an example of a rational function. Remember, a rational function is a quotient between two polynomials. And this is a useful result because we know we can use direct substitution to evaluate the limit of a rational function as long as the denominator does not evaluate to give us zero. And in this case, our limit is as 𝑥 is approaching negative five. And if we substitute 𝑥 is equal to negative five into 𝑥 plus seven, we get two. So this is not zero. This means we can evaluate the limit of just the first term in our numerator one divided by 𝑥 plus seven by using direct substitution.
We can ask the same question about the second term in our numerator negative one-seventh. This is a constant function, so we can evaluate this by using direct substitution. Alternatively, you can think of this as a polynomial with degree zero. In fact, it might be a little bit strange to say that we’re evaluating the limit of a constant by using direct substitution since the limit of a constant will always just be equal to that constant itself. However, it does work. And we can technically do it this way. It’s just that our constant is the same regardless of what value of 𝑥 we substitute in.
We now have one last thing we need to check. We’re dividing through by 𝑥. And we know that 𝑥 is an example of a polynomial. And we also know that we can use direct substitution to evaluate the limits of polynomials. So in our case, our limit is the sum, difference, product, and quotient of functions which we can evaluate by using direct substitution. And we know a useful property about functions of this form. We know if 𝑓 of 𝑥 is the sum, difference, product, or quotient of functions which we know we can use direct substitution to evaluate their limit, then as long as we’re not dividing by zero, we can just evaluate the limit of this function by using direct substitution.
And this is true in this case. We can see we substitute 𝑥 is equal to negative five into our function. We will never be dividing by zero. Now, we’re ready to evaluate this limit by direct substitution. We just substitute in 𝑥 is equal to negative five. This gives us one divided by negative five plus seven minus one-seventh all divided by negative five. And if we evaluate this, we get negative one divided by 14. And this is our final answer.
It’s also worth pointing out there is one more method we could’ve used to answer this question. We could’ve used algebraic manipulation to rewrite the function inside of our limit as a rational function. Then we would just use our rules for evaluating the limit of rational functions by using direct substitution. This would also work and is a perfectly valid way of answering this question. However, in this instance, we were able to show that we could evaluate this limit by using direct substitution by showing it was comprised of a rational function and polynomials. Therefore, we could evaluate this limit by just substituting in 𝑥 is equal to negative five. And we got that this is evaluated our limit at negative one divided by 14.