Question

How do you evaluate the limit `(7x^2-4x-3)/(3x^2-4x+1)` as x approaches 1?

Answer 1

`lim_(x->1) (7x^2-4x-3)/(3x^2-4x+1) = 5`

Explanation:

Factoring the numerator and denominator, we can simplify the rational expression as follows:

`(7x^2-4x-3)/(3x^2-4x+1) = (color(red)(cancel(color(black)((x-1))))(7x+3))/(color(red)(cancel(color(black)((x-1))))(3x-1)) = (7x+3)/(3x-1)`

with exclusion `x != 1`

That means that the functions `(7x^2-4x-3)/(3x^2-4x+1)` and `(7x+3)/(3x-1)` are identical, except that the first one has a hole at `x=0`, while the second one does not.

So we find:

`lim_(x->1) (7x^2-4x-3)/(3x^2-4x+1) = lim_(x->1) (7x+3)/(3x-1)`

`color(white)(lim_(x->1) (7x^2-4x-3)/(3x^2-4x+1)) = (7(color(blue)(1))+3)/(3(color(blue)(1))-1)`

`color(white)(lim_(x->1) (7x^2-4x-3)/(3x^2-4x+1)) = (7+3)/(3-1)`

`color(white)(lim_(x->1) (7x^2-4x-3)/(3x^2-4x+1)) = 10/2`

`color(white)(lim_(x->1) (7x^2-4x-3)/(3x^2-4x+1)) = 5`