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

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

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Find lim_(π₯ β 2) (8π₯Β³ β 64)/(π₯Β² β 4).

02:27

### Video Transcript

Find the limit as π₯ approaches two of eight π₯ cubed minus 64 divided by π₯ squared minus four.

Here, we have a function, which weβll call π of π₯. Given that this is a rational function, the first thing we may try is a direct substitution of π₯ equals two into our function. Here, we have performed the substitution. And when we evaluate our answer, we find that weβre left with the indeterminate form of zero over zero. Instead, weβre gonna need to move on to a different method based on factorisation.

For this method, we first note 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. By inspecting the numerator of our quotient, we noticed that both the terms have a factor of eight. We can, therefore, factorise our numerator as eight times π₯ cubed minus eight. And given that this eight is a constant, we can take it outside of our limit as follows. If instead we find the limit as π₯ approaches two of π₯ cubed minus eight divided by π₯ squared minus four and multiply this entire thing by our constant eight, this will give us the same answer.

To proceed, we can then notice that the eight in our numerator and the four in our denominator can both be expressed as powers of two, which are two cubed and two squared, respectively. After doing this, we see that our limit now takes the following form. Here, weβll say that the top half of our quotient is equal to the difference of two πth powers, with π being three, and the bottom half of our quotient is equal to the difference of two πth powers, with π being two.

Given this form, weβre able to use the following general rule, which tells us that the limit will be equal to π over π times π to the power of π minus π. At this point, you may notice that both the top and bottom half of our quotient would have a common factor of π₯ minus two. We could, instead, cancel this common factor and proceeded by refactorising. However, this general rule allows us to move directly to our limit.

In cases, where π or π are large, this helps us potentially avoid a lengthy or time-consuming refactorisation. Inputting our values into our general rule, where π is two, π is three, and π is also two, we find that our limit is three over two multiplied by two to the power of three minus two. We also mustnβt forget to multiply this entire thing by the eight which we took out of our limit.

Three minus two is, of course, just one. And so, we can simplify this by cancelling the two and the one over two. We then find that our answer is equal to eight times three, which 24. We have now found that the limit, as π₯ approaches two of our function π of π₯, is equal to 24. And we have answered our question.

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