Video: Finding an Unknown in a Rational Function given Its Limit at a Point

Given that lim_(π‘₯ β†’ βˆ’5) π‘Ž/(π‘₯ βˆ’ 1) = 6, what is π‘Ž?

02:55

Video Transcript

Given that the limit as π‘₯ approaches negative five of π‘Ž divided by π‘₯ minus one is equal to six, what is π‘Ž?

The question gives us the limit and tells us that the evaluation of this limit is equal to six. It wants us to calculate the value of π‘Ž. We can do this by attempting to evaluate this limit ourselves and then remembering that this must be equal to six. The first thing we should try using if we want to evaluate this limit is direct substitution. We see that π‘Ž divided by π‘₯ minus one is a rational function. This is because it’s the quotient of two polynomials, 𝑝 of π‘₯ and π‘ž of π‘₯.

And we know that the constant π‘Ž is a polynomial. And we know if we have a rational function, 𝑓 of π‘₯, and the polynomial in this denominator evaluated at 𝑏 is not equal to zero, then we can evaluate this limit by using direct substitution. We’ll have the limit as π‘₯ approaches 𝑏 of our rational function 𝑓 of π‘₯ is equal to 𝑓 evaluated at 𝑏. In fact, even in some cases where our denominator evaluates to give us zero, we can check if the polynomial in our numerator also evaluates to give us zero.

Then, we can use the factor theorem to take out a factor of π‘₯ minus 𝑏 from both our numerator and our denominator, and then cancel the shared factor of π‘₯ minus 𝑏. We want the limit as π‘₯ is approaching negative five, so we’ll set 𝑏 equal to negative five. And we’re calculating the limit of π‘Ž divided by π‘₯ minus one. So, we’ll set the polynomial in our numerator 𝑝 of π‘₯ to be equal to π‘Ž and the polynomial in our denominator π‘ž of π‘₯ to be equal to π‘₯ minus one. So, for our rational function, 𝑓 of π‘₯ is equal to π‘Ž divided by π‘₯ minus one.

To see if we can use direct substitution, we need to check if π‘₯ minus one evaluated at negative five is equal to zero. So, we evaluate our denominator, π‘ž of π‘₯, at π‘₯ is equal to negative five. This gives us negative five minus one, which is equal to negative six. So, we’ve shown that our denominator evaluated at π‘₯ is equal to negative five is not equal to zero. So, we’re allowed to use direct substitution. So, by using direct substitution, we have the limit in our question, the limit as π‘₯ approaches negative five of π‘Ž divided by π‘₯ minus one is equal to π‘Ž divided by negative five minus one. And this simplifies to give us π‘Ž divided by negative six.

However, the question tells us that this limit is also equal to six. So, we have six is equal to our limit. And we’ve already shown that by direct substitution, our limit is equal to π‘Ž divided by negative six. So, we must have six is equal to negative π‘Ž divided by six. And we can solve this by multiplying both sides of our equation by negative six. This gives us negative six times six, which is equal to negative 36, is equal to π‘Ž. Therefore, we have shown if the limit as π‘₯ approaches negative five of π‘Ž divided by π‘₯ minus one is equal to six, then the value of π‘Ž must be equal to negative 36.

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