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

Find lim_(π‘₯ β†’ βˆ’8) (3π‘₯ + 24)/(5π‘₯Β² + 34π‘₯ βˆ’ 48).

04:11

Video Transcript

Find the limit as π‘₯ approaches negative eight of three π‘₯ plus 24 all divided by five π‘₯ squared plus 34π‘₯ minus 48.

We see that the question wants us to evaluate the limit of a rational function. That’s the quotient of two polynomials. The first thing we can try is to evaluate this by direct substitution. We recall we can use direct substitution to evaluate the limit of a rational function if the polynomial in the denominator of our rational function evaluated at π‘Ž is not equal to zero. This would then give us the limit as π‘₯ approaches π‘Ž of our rational function 𝑓 of π‘₯ is equal to 𝑓 evaluated at π‘Ž.

We want to find the limit as π‘₯ approaches negative eight. So we’ll set π‘Ž equal to negative eight. We’ll set 𝑃 of π‘₯ equal to the polynomial in the numerator of our limit. That’s three π‘₯ plus 24. And we’ll set π‘ž of π‘₯ equal to the polynomial in the denominator of our limit. That’s five π‘₯ squared plus 34π‘₯ minus 48. So we need to check π‘ž evaluated at negative eight.

We have that our denominator π‘ž evaluated at negative eight is five times negative eight squared plus 34 times negative eight minus 48. Simplifying this, we get 320 minus 272 minus 48, which we can calculate to be equal to zero. So our denominator evaluated at negative eight is equal to zero. So negative eight is not in the domain of our function 𝑓 of π‘₯. So we can’t use direct substitution.

We now need to check what happens to our numerator when π‘₯ is equal to negative eight. This will help us determine if the limit exists or if the limit does not exist. We see that our numerator 𝑃 evaluated at negative eight is equal to three times negative eight plus 24, which we can calculate to give us zero. Since 𝑃 evaluated at negative eight and π‘ž evaluated at negative eight are both equal to zero, by the factor theorem we have that π‘₯ plus eight must be a factor of both 𝑃 of π‘₯ and π‘ž of π‘₯. So both our numerator and our denominator in our function 𝑓 of π‘₯ share a factor of π‘₯ plus eight.

We can take out the shared factor of three in our numerator to get three multiplied by π‘₯ plus eight. And we know that π‘₯ plus eight divides the quadratic in our denominator by the factor theorem. Since the coefficient of π‘₯ squared in our quadratic is five, our second factor must have a leading term of five π‘₯. Similarly, we see that eight multiplied by negative six is equal to negative 48.

If we were to then cancel the shared factor of π‘₯ plus eight in our numerator and our denominator, we get three divided by five π‘₯ minus six. And it’s worth noting that three divided by five π‘₯ minus six is equal to our function 𝑓 of π‘₯ everywhere except when π‘₯ is equal to negative eight. And this is because our function 𝑓 of π‘₯ was undefined when π‘₯ was equal to negative eight. It gave us zero divided by zero, an indeterminate form.

But remember, when we’re calculating the limit as π‘₯ approaches a value, we don’t care what happens when π‘₯ is equal to that value. We only care what happens when π‘₯ is near that value. In other words, if we have two functions, 𝑓 of π‘₯ and 𝑔 of π‘₯ which are equal around our value, π‘₯ is equal to π‘Ž, and we know the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ is equal to 𝐿. Then we can conclude the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is also equal to 𝐿.

So because we have that our function 𝑓 of π‘₯ is equal to three divided by five π‘₯ minus six everywhere except when π‘₯ is equal to negative eight. We have the limit as π‘₯ approaches negative eight of our function 𝑓 of π‘₯ is equal to the limit as π‘₯ approaches negative eight of three divided by five π‘₯ minus six.

In fact, we see that this is a rational function and its denominator evaluated at negative eight is not equal to zero. So we can evaluate this using direct substitution. So by direct substitution, we have that our limit is equal to three divided by five multiplied by negative eight minus six, which we can calculate to give us negative three divided by 46.

Therefore, we’ve shown that the limit as π‘₯ approaches negative eight of three π‘₯ plus 24 all divided by five π‘₯ squared plus 34π‘₯ minus 48 is equal to negative three divided by 46.

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