Video: Finding the Limit of a Rational Function at a Point

Determine lim_(π‘₯ β†’ βˆ’4) ((π‘₯Β² βˆ’ 9π‘₯ + 2)/(βˆ’8π‘₯Β² βˆ’ 3π‘₯ βˆ’ 9)).

03:11

Video Transcript

Determine the limit as π‘₯ approaches negative four of π‘₯ squared minus nine π‘₯ plus two all divided by negative eight π‘₯ squared minus three π‘₯ minus nine.

When asked to evaluate a limit, the first thing we should try is direct substitution. We can notice that we’re asked to evaluate the limit of a rational function. That’s the quotient of two polynomials, 𝑝 of π‘₯ divided by π‘ž of π‘₯. And we recall that we can evaluate the limit as π‘₯ approaches some finite value of π‘Ž of a rational function by using direct substitution if the polynomial in the denominator evaluated at π‘Ž does not give us zero. This gives us that the limit as π‘₯ approaches π‘Ž of our function 𝑓 of π‘₯ is equal to 𝑓 evaluated at π‘Ž.

In fact, even if the polynomial in our denominator evaluated at π‘Ž was equal to zero, we could check if the polynomial in our numerator evaluated at π‘Ž was also equal to zero. We could then apply the factor theorem to note that both of our polynomials share a factor of π‘₯ minus π‘Ž. Cancelling this shared factor of π‘₯ minus π‘Ž would then give us a new rational function, which agrees everywhere with our old rational function 𝑓 of π‘₯ except to the point where π‘₯ is equal to π‘Ž. We could then attempt to use direct substitution on this new rational function to evaluate our limit.

We want to calculate the limit as π‘₯ approaches negative four. So, we’ll set π‘Ž equal to negative four. And we’ll set 𝑝 of π‘₯ equal to the polynomial in the numerator of our limit. That’s π‘₯ squared minus nine π‘₯ plus two. And we’ll set π‘ž of π‘₯ equal to the polynomial in the denominator of our limit. That’s negative eight π‘₯ squared minus three π‘₯ minus nine.

We now need to check the output of our denominator when we input π‘₯ is equal to negative four. Substituting π‘₯ is equal to negative four into our polynomial π‘ž of π‘₯ gives us negative eight times negative four squared minus three times negative four minus nine. And we can simplify this to give us negative 128 plus 12 minus nine, which is equal to negative 125. And, in particular, this is not equal to zero. So, we can use direct substitution to evaluate the limit given to us in the question.

So by using direct substitution, we have the limit as π‘₯ approaches negative four of π‘₯ squared minus nine π‘₯ plus two all divided by negative eight π‘₯ squared minus three π‘₯ minus nine is equal to negative four squared minus nine times negative four plus two all divided by negative eight times negative four squared minus three times negative four minus nine. We can simplify this calculation since the denominator is equal to π‘ž evaluated at negative four, which we’ve already found to be equal to negative 125.

So, we now only need to calculate our numerator. We have negative four squared is equal to 16. And then, we subtract nine times negative four which is positive 36. And then, we add two. And we can simplify this to give us negative 54 divided by 125.

Therefore, we’ve shown the limit as π‘₯ approaches negative four of π‘₯ squared minus nine π‘₯ plus two all divided by negative eight π‘₯ squared minus three π‘₯ minus nine is equal to negative 54 divided by 125.

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