Question

How do you evaluate `lim_(x->3)(2x^2-5x-3)/(x-3)`?

Answer 1

`lim_(xrarr3)(2x^2-5x-3)/(x-3) = lim_(xrarr3)((x-3)(2x+1))/(x-3) = lim_(xrarr3)(2x+1) = 2(3)+1 = 7`

Explanation:

The numerator is `0` when `x=3` (`3` is a zero of the numerator).

Therefore `x-3` is a factor of the numerator. (So the quotient can be simplified for `x != 3`)

`2x^2-5x-3 = (x-3)( "something")`

By trial and error (or no error, just trial):

`2x^2-5x-3 = (x-3)(2x+1)`

So `(2x^2-5x-3)/(x-3) = ((x-3)(2x+1))/(x-3) = 2x+1` for `x != 3`.

The limit as `xrarr3` doesn't care what happens right at `3`, just for `x` close to `3`, so we can find the limit.

`lim_(xrarr3)(2x^2-5x-3)/(x-3) = lim_(xrarr3)((x-3)(2x+1))/(x-3) = lim_(xrarr3)(2x+1) = 2(3)+1 = 7`

Answer 2

The function approaches 7

Explanation:

To find the limit of a function, we use L'Hopital's rule, where: `lim_(x->a)f(x)/g(x)=(f'(x))/(g'(x)), a!=\oo` then substitute `a` for `x`.

So, for our example: `lim_(x->3)(2x^2-5x-3)/(x-3)`, to find the limit we differentiate both equations to get:
`lim_(x->3)(4x-5)/1`

Now we replace `x` with 3:
`lim_(x->3)(4(3)-5)/1=7`