Question Video: Finding the Limit of a Rational Function at a Point | Nagwa Question Video: Finding the Limit of a Rational Function at a Point | Nagwa

# Question Video: Finding the Limit of a Rational Function at a Point Mathematics • Second Year of Secondary School

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Find lim_(π₯ β β8) (π₯Β² + 13π₯ + 40)/(π₯Β³ + 9π₯Β² β 12π₯ β 160).

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### Video Transcript

Find the limit as π₯ approaches negative eight of π₯ squared plus 13π₯ plus 40 divided by π₯ cubed plus nine π₯ squared minus 12π₯ minus 160.

Here, we see that we are trying to find the limit of a rational function, which weβll call π of π₯. For this type of function, if negative eight is in the domain of π, then the limit, as π₯ approaches negative eight, is simply π evaluated at negative eight. The first thing we can do then is to try direct substitution of negative eight into our function.

Here, we have performed the substitution. And following through with our working, we find that our answer evaluates to zero over zero. And this is an indeterminate form, which means we cannot evaluate the limit using direct substitution. We can also conclude that negative eight is not in the domain of our function π. Here, weβre gonna need to use a different approach. And we can do so first by recognising that our function π of π₯ is in the form π of π₯ over π of π₯, where both π and π are polynomial functions.

Looking at the failed direct substitution that we just tried, we can see the both π of negative eight and π of negative eight are equal to zero. From this information, we can use the factor theorem to conclude that π₯ plus eight is a factor of both π of π₯ and π of π₯. Since direct substitution failed, letβs instead try to factorise both the top and bottom halves of our quotient, given the fact that both of them have a common factor of π₯ plus eight.

For the numerator, factoring a quadratic equation should be a familiar skill to us. And with some inspection, we see that π₯ squared plus 13π₯ plus 40 factorises to π₯ plus eight and π₯ plus five. Now, for the denominator, ordinarily, factoring a cubic is a much more difficult task. However, given the fact that we already know one of the factors is π₯ plus eight, we can use techniques such as polynomial division or comparing coefficients to find the other factor.

For this video, weβre going to compare coefficients. And we begin by recognising that our other factor will take the form ππ₯ squared plus ππ₯ plus π. To find π, we note that we have an π₯ multiplied by an ππ₯ squared. And this is equal to π₯ cubed. It, therefore, follows that π is equal to one. And we then see that our second factor begins with the term π₯ squared.

Next, to find π, we note that we have eight multiplied by π is equal to negative 160. And we can calculate that π is equal to negative 20. Finally, to find π, weβll choose to look at the nine π₯ squared term. Going by the previous coefficients we have, we note that we first have an eight multiplied by an π₯ squared. This gives us eight π₯ squared. We also have an π₯ multiplied by ππ₯ and this gives us ππ₯ squared. The sum of these two terms is nine π₯ squared. And it then follows that π is equal to one.

The denominator of our quotient is now fully factorised. And our missing factor was π₯ squared plus π₯ minus 20. Following this factorisation, weβre able to cancel the common factor of π₯ plus eight from the top and bottom half of our quotient. And we are then left with π₯ plus five over π₯ squared plus π₯ minus 20.

Now, here, we should remember that in cancelling the common factor of π₯ plus eight, we have changed the domain of our original function π of π₯. Weβre able to say that π of π₯ is equal to the right-hand side of our equation, which weβll call π of π₯, at all values where π₯ is not equal to negative eight. It now follows that the limit, as π₯ approaches negative eight of π of π₯, is equal to the limit as π₯ approaches negative π₯ of π of π₯. Crucially, this is because the limit concerns values where π₯ is close to negative eight but not equal to negative eight.

An important point is that π of π₯ is defined when π₯ is equal to negative eight. And so, we can, therefore, find the limit by direct substitution. Here, we have performed the substitution of π of negative eight. And if we follow through with our calculations, we see that this evaluates to negative three over 36. This fraction can be simplified to negative one over 12. We have now answered the question. And we have found our limit.

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