Question

How do you find the limit of `(3x^2-x-10)/(x^2+5x-14)` as x approaches 2?

Answer 1

`11/9`

Explanation:

Observe that
`3x^2-x-10=(3x+5)(x-2)`
`x^2+5x-14=(x+7)(x-2)`
So we get

`lim_(x to 2)(3x^2-x-10)/(x^2+5x-1)=lim_(x to 2)(3x+5)/(x+7)=11/9`

Answer 2

`lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = 11/9`

Explanation:

Method 1): factorize the polynomials

As both numerator and denominator vanish for `x=2` they can be divided by `(x-2)`:

`3x^2-x-10 = (3x+5)(x-2)`

`x^2+5x-14 = (x+7)(x-2)`

Then:

`lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) ( (3x+5)(x-2))/((x+7)(x-2))`

`lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) (3x+5)/(x+7)`

`lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = 11/9`

Method 2): L'Hospital's rule:

As both numerator and denominator vanish for `x=2`:

`lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) (d/dx (3x^2-x-10))/(d/dx (x^2+5x-14 ) )`

`lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) (6x-1)/(2x+5) = 11/9`