Question Video: Finding Unknowns in a Rational Function given Its Limit at a Point | Nagwa Question Video: Finding Unknowns in a Rational Function given Its Limit at a Point | Nagwa

Question Video: Finding Unknowns in a Rational Function given Its Limit at a Point Mathematics • Higher Education

Given that lim_(π‘₯ β†’ βˆ’1) (π‘₯Β² βˆ’ (π‘š βˆ’ 1)π‘₯ βˆ’ π‘š)/(π‘₯ + 1) = βˆ’3, determine the value of π‘š.

05:05

Video Transcript

Given that the limit as π‘₯ approaches negative one of π‘₯ squared minus π‘š minus one times π‘₯ minus π‘š all divided by π‘₯ plus one is equal to negative three, determine the value of π‘š.

The question tells us that the limit as π‘₯ approaches negative one of a rational function is equal to negative three. It wants us to use this information to determine the value of π‘š. Since this is the limit of a rational function, we can attempt to evaluate this by using direct substitution. Substituting π‘₯ is equal to negative one in the function in our limit gives us negative one squared minus π‘š minus one times negative one minus π‘š all divided by negative one plus one. Fully evaluating the expressions in our numerator and our denominator, we see we get the indeterminate form zero divided by zero. So evaluating our limit by direct substitution gained us no information about the value of π‘š. So we’re going to need a different way to evaluate this limit.

We recall that L’HΓ΄pital’s rule gives us a way of evaluating an indeterminate limit of this form. We recall the following version of L’HΓ΄pital’s rule. If we have two differentiable functions 𝑓 and 𝑔 where 𝑔 prime of π‘₯ is not equal to zero around our value of π‘Ž except possibly when π‘₯ is equal to π‘Ž, and the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ and the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ are both equal to zero. Then the limit as π‘₯ approaches π‘Ž of the quotient of 𝑓 of π‘₯ and 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of the quotient of 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. In this version of L’HΓ΄pital’s rule, we’re told that if the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ gives us the indeterminate form of zero over zero. Then, under these conditions, we can instead calculate the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯.

Let’s start by verifying that we can actually use L’HΓ΄pital’s rule in this case. We’ll set our function 𝑓 of π‘₯ equal to the numerator of the function in our limit. That’s π‘₯ squared minus π‘š minus one times π‘₯ minus π‘š. And we’ll set our function 𝑔 of π‘₯ to be the function in the denominator of our limit. That’s 𝑔 of π‘₯ is equal to π‘₯ plus one. The first thing we need to use this version of L’HΓ΄pital’s rule is that both our functions 𝑓 and 𝑔 are differentiable. We see the 𝑔 is a linear function, and for any value π‘š, 𝑓 of π‘₯ is a quadratic. So both of these functions are polynomials, and therefore they’re both differentiable.

Next, we need to show that derivative of 𝑔 of π‘₯ is not equal to zero around π‘Ž. Remember, π‘Ž is the constant that π‘₯ is approaching in our limit. We have the limit as π‘₯ approaches negative one. So we’ll set π‘Ž equal to negative one. Since 𝑔 of π‘₯ is the linear function π‘₯ plus one, we can just calculate 𝑔 prime of π‘₯ directly by using the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. In this case, we just get the constant one. And we can, of course, see that this is never equal zero.

Next, we need to show that the limit as π‘₯ approaches negative one of 𝑓 of π‘₯ and the limit as π‘₯ approaches negative one of 𝑔 of π‘₯ is both equal to zero. However, we’ve actually already shown this. When we tried to use direct substitution to evaluate our limit, we saw that our numerator evaluated to give us zero and our denominator also evaluated to give us zero. But the numerator was just the limit as π‘₯ approaches negative one of 𝑓 of π‘₯. And the denominator was the limit as π‘₯ approaches negative one of 𝑔 of π‘₯. So we’ve shown the last two prerequisites of this version of the L’HΓ΄pital’s rule is true. So we can attempt to use it to evaluate our limit.

So by using this version of L’HΓ΄pital’s rule, we have the limit given to us in the question, which is the limit as π‘₯ approaches negative one of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches negative one of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯. So to evaluate this limit, we need to find 𝑓 prime of π‘₯. We recall 𝑓 of π‘₯ is equal to π‘₯ squared minus π‘š minus one times π‘₯ minus π‘š. And for any value of π‘š, this is just a quadratic. So we can evaluate this derivative by using the power rule for differentiation. We get two π‘₯ minus π‘š minus one. So now that we found 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯, we’ve shown that our limit is equal to the limit as π‘₯ approaches negative one of two π‘₯ minus π‘š minus one divided by one. Of course, dividing by one inside of our limit doesn’t change anything.

And now we see we’re just calculating the limit of a linear function. We can evaluate this by using direct substitution. We substitute π‘₯ is equal to negative one into our linear function to evaluate the limit. And substituting π‘₯ is equal to negative one into this limit gives us two times negative one minus π‘š minus one. And if we evaluate and simplify this expression, we can see that it’s equal to negative one minus π‘š. So we’ve shown that the limit given to us in the question is equal to negative one minus π‘š. But the question also tells us that this limit must be equal to negative three. So, negative one minus π‘š is equal to negative three. And we can then solve this equation. We get that π‘š is equal to two.

So we’ve shown if the limit as π‘₯ approaches negative one of π‘₯ squared minus π‘š minus one times π‘₯ minus π‘š all divided by π‘₯ plus one is equal to negative three, then the value of π‘š must be equal to two.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy