`lim_(x->3) (x^3+7x^2-32x+6)/(x^3-9x)`

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We can use the Rational Root Theorem and Synthetic Division to factor the top.

The theorem mention about tells us to take the ratio of the constant term (in this case 6), and the leading coefficient (in this case 1 from x^3). The ratio will be the possible zeros of our polynomial on top.

The possible zeros will be: `+-6, +-3, +-2,+-1.`

Let us try to use x = 3.

3| 1 7 -32 6

+ 3 30 -6

_____________________

1 10 -2 0

Since, the remainder is zero, the x = 3 is a zero of our polynomial.

So, the factored form of the top is `(x - 3)(x^2 + 10x -2)`

For the bottom, factor out an x first. ` `

`x^3 - 9x = x(x^2 - 9)`

Use difference of two squares for x^2 - 9.

`x(x^2 - 9) = x(x - 3)(x + 3)`

So, we will have:

`((x - 3)(x^2+ 10x -2))/(x(x - 3)(x + 3))`

Cancel common factor on top and bottom.

`(x^2 + 10x - 2)/(x(x + 3))`

We can now take the limit as x approaches 3.

`lim_(x->3) (x^2 + 10x - 2)/(x(x + 3)) = (3^2 + 10(3) - 2)/(3(3 + 3)) = (9 + 30 - 2)/(3 * 6) = 37/18`

Hence, the **limit as x-> 3 is 37/18**.

Since the limit is indeterminate, type `0/0` , you may also evaluate it using l'Hospital's theorem, such that:

`lim_(x->3) (x^3 + 7x^2 - 32x + 6)/(x^3 - 9x)= lim_(x->3) ((x^3 + 7x^2 - 32x + 6)')/((x^3 - 9x)')`

`lim_(x->3) ((x^3 + 7x^2 - 32x + 6)')/((x^3 - 9x)') = lim_(x->3) (3x^2 + 14x - 32)/(3x^2 - 9)`

You need to check if the limit is indeterminate, replacing 3 for x such that:

`lim_(x->3) (3x^2 + 14x - 32)/(3x^2 - 9) = (3*3^2 + 14*3 - 32)/(3*3^2 - 9) `

`lim_(x->3) (3x^2 + 14x - 32)/(3x^2 - 9) = (27 + 42 - 32)/(27 - 9)`

`lim_(x->3) (3x^2 + 14x - 32)/(3x^2 - 9) = 37/18`

Hence, evaluating the limit using l'Hospital's theorem, yields `lim_(x->3) (x^3 + 7x^2 - 32x + 6)/(x^3 - 9x)= 37/18.`

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