# What is `lim_(x->4) (x^3-2x^2-9x+4)/(x^2-2x-8)`

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### 2 Answers

The value of `lim_(x->4) (x^3-2x^2-9x+4)/(x^2-2x-8)` has to be determined.

`lim_(x->4) (x^3-2x^2-9x+4)/(x^2-2x-8)`

=> `lim_(x->4) ((x-4)*(x^2+2x-1))/((x-4)*(x+2))`

=> `lim_(x->4) (x^2+2x-1)/(x+2)`

substituting x = 4 gives

`(16 + 8 - 1)/6`

=> `23/6`

**The limit `lim_(x->4) (x^3-2x^2-9x+4)/(x^2-2x-8) = 23/6`**

### User Comments

The limit `lim_(x->4)(x^3 - 2x^2 - 9x + 4)/(x^2 - 2x - 8)` is required.

The fist step in determining a limit is to substitute the value that the variable tends to in the given expression. If the result of doing this is a rational number that is the required limit.

In the given problem, if we substitute x = 4 the result is:

`(4^3 - 2*4^2 - 9*4 + 4)/(4^2 - 2*4 - 8)`

= `(64 - 2*16 - 36 + 4)/(16 - 8 - 8)`

= `0/0`

This is an indeterminate form and we can use l'Hospital's rule and replace the numerator and denominator by their derivatives.

This gives:

`lim_(x->4) (3x^2 - 4x - 9)/(2x - 2)`

Substituting x = 4 gives:

`(3*16 - 16 - 9)/(8 - 2)`

= `23/6`

The required limit `lim_(x->4)(x^3 - 2x^2 - 9x + 4)/(x^2 - 2x - 8) = 23/6`