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 Question from selvamani, a student: F (x) = x^3+3x^2-9x-2 / x^3-x-6 and limit x tends to 2, f (x) exist then limit x tends to 2, f (x) is equal to ? How to answer this problem.

Hi Selvamani,

Both the numerator $x^3+3x^2-9x-2$ and the denominator $x^3-x-6$ are zero when $x = 2$ and hence you ca use the factor theorem on both of them. For example, since the denominator is zero when $x = 2$ there is a polynomial $p(x)$ so that $x^3+3x^2-9x-2 = (x - 2) p(x).$ You can find $p(x)$ by dividing $x^3+3x^2-9x-2$ by $x - 2.$ Similarly for the denominator there is a polynomial $q(x)$ so that $x^3-x-6 = (x - 2) q(x).$ Thus

$\frac{x^3+3x^2-9x-2}{x^3-x-6} = \frac{(x - 2) p(x)}{(x - 2) q(x)}$

and since $x \neq 2$ the $(x - 2)$ factors can be cancelled to get

$\frac{x^3+3x^2-9x-2}{x^3-x-6} = \frac{p(x)}{q(x)}.$

Penny

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