### Video Transcript

Find the limit as π₯ approaches negative seven of π₯ squared minus 49 divided by three π₯ cubed plus 24π₯ squared plus 13π₯ minus 56.

We see the question wants us to evaluate the limit of a rational function. Thatβs the quotient of two polynomials. And we know we can attempt to evaluate the limit of a rational function by using direct substitution. So, letβs try to evaluate this limit by using direct substitution. Weβll substitute π₯ is equal to negative seven into our rational function. This gives us negative seven squared minus 49 divided by three times negative seven cubed plus 24 times negative seven squared plus 13 times negative seven minus 56.

Simplifying the expressions in our numerator and our denominator, we get 49 minus 49 divided by negative 1029 plus 1176 minus 91 minus 56. And if we evaluate this expression, we get zero divided by zero, which is an indeterminate form. This means we cannot directly evaluate this limit by using direct substitution. It does not mean we canβt evaluate this limit at all; it just means we canβt do it directly by using direct substitution. Weβre going to need to find a different method.

When we substituted π₯ is equal to negative seven into our numerator and into our denominator, we got zero for both of these polynomials. So, we can actually apply the factor theorem to both the polynomial in our numerator and our denominator. We get that π₯ plus seven is a factor of both of these polynomials. So, letβs clear some space and use the factor theorem to take out a factor of π₯ plus seven from the polynomial in our numerator and the polynomial in our denominator.

If we look at our numerator, we see itβs of the form π₯ squared minus 49, and 49 is seven squared. And we recall we can factor the difference of squares, π₯ squared minus π squared, as π₯ plus π multiplied by π₯ minus π. So, we can fully factor the numerator as π₯ plus seven times π₯ minus seven. We now want to take out our factor of π₯ plus seven from the polynomial in our denominator. Thereβs a few different ways of doing this. For example, we could use polynomial division. Another way of doing this is to notice that a linear function multiplied by a quadratic will give us a cubic.

So, we have π₯ plus seven times some quadratic weβve labeled ππ₯ squared plus ππ₯ plus π is equal to the cubic in the denominator of our limit. And we can find the values of π, π, and π by equating the coefficients of the powers of π₯. The coefficient of π₯ cubed in our cubic is three. And when we multiply our linear factor and our quadratic factor, we see that the coefficient of π₯ cubed will be π. So because the coefficients of π₯ cubed must be equal, we have π is equal to three.

We can do something similar to find the value of π. We notice the constant in our cubic is equal to negative 56. And when we multiply our linear factor and our quadratic factor, we see that our constant will be equal to seven times π. So, by equating the constant terms, we have negative 56 is equal to seven π. And dividing through by seven, we see that π is equal to negative eight.

Finding the value of π is slightly more difficult. Weβll do this by equating the coefficients of π₯. In our cubic, the coefficient of π₯ is 13. And when we multiply our linear factor and our quadratic factor, we see that there are two ways of getting coefficients of π₯. So, we need to add these two together. This gives us that our coefficient of π₯ will be seven π plus one times negative eight. And this will be equal to the coefficient of π₯ in our cubic, which is 13. And we can solve this equation for π. We get that π is equal to three.

So, weβve shown that our limit is equal to the limit as π₯ approaches negative seven of π₯ plus seven times π₯ minus seven divided by π₯ plus seven times three π₯ squared plus three π₯ minus eight. And we can now see that both our numerator and our denominator share a factor of π₯ plus seven. We want to cancel this shared factor. And itβs worth thinking about the question, are we allowed to do this?

We want to just cancel this shared factor to show that our limit is equal to the limit as π₯ approaches negative seven of π₯ minus seven divided by three π₯ squared plus three π₯ minus eight. But are these two limits actually equal? We notice that when π₯ is not equal to negative seven, then π₯ plus seven is not equal to zero. So, if π₯ plus seven is not equal to zero, in this case, we can just cancel the shared factors.

So, the only problem we have is when π₯ is equal to negative seven. But remember, weβre only interested in the limit as π₯ approaches negative seven. And when weβre evaluating the limit of a function as π₯ approaches negative seven, weβre not actually interested in what happens at the function when π₯ is equal to negative seven. Weβre only interested in what happens to our function as π₯ approaches negative seven. So, our new function is exactly equal to the function we had before everywhere except when π₯ is equal to negative seven. So, their limits as π₯ approaches negative seven will be equal. So, this justifies us canceling the shared factor of π₯ plus seven.

We now see, we want to evaluate the limit of this new rational function. We can attempt to do this by direct substitution. Substituting π₯ is equal to negative seven, we get negative seven minus seven divided by three times negative seven squared plus three times negative seven minus eight. And we can evaluate this quotient to get negative seven divided by 59.

So, weβve shown the limit as π₯ approaches negative seven of π₯ squared minus 49 divided by three π₯ cubed plus 24π₯ squared plus 13π₯ minus 56 is equal to negative seven divided by 59.