Question

How do you evaluate `[(x^2) + 5x + 4] / [(x^2) + 3x -4]` as x approaches -4?

Answer 1

Hence the limit is an undefined form `0/0` we can apply L'Hopital law so we have that

#lim_(x->-4) [x^2+5x+4]/[x^2+3x+4]= lim_(x->-4) (d[x^2+5x+4]/dx)/(d[x^2+3x+4]/dx)= lim_(x->-4) (2x+5)/(2x+3)=(2*(-4)+5)/(2*(-4)+3)= (-3)/(-5)=3/5#

Answer 2

If you have not yet learned about derivatives and l'Hospital's rule, you can still find the limit.

Explanation:

`lim_(xrarr-4)(x^2+5x+4)/(x^2+3x-4)` has indeterminate initial form. `-4` is a zero of both the numerator and denominator.

The numerator and denominator are polynomials, and we know from Algebra, that if `z` is a zero of a polynomial, then `x-z` is a factor of the polynomial.

So we know that `x-(-4) = x+4` is a factor of both the numerator and the denominator of `f`.

So we can factor and reduce and try again to evaluate the limit.
Factoring is simplified by the fact that we already know one of the factors is `x+4`

`(x^2+5x+4)/(x^2+3x-4) = ((x+4)(x+1))/((x+4)(x-1))`.

Now for every `x` other than `-4`,

`((x+4)(x+1))/((x+4)(x-1)) = (x+1)/(x-1)`

(for `x=-4` there is no number of the left, so there is nothing there to be equal or different from anything.)

The limit as `x` approaches `-4` doesn't care what happens when `x` actually equals `-4`, so

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

`= lim_(xrarr-4)(x+1)/(x-1)`

We can evaluate this limit by substitution. We get

`lim_(xrarr-4)(x+1)/(x-1) = ((-4)+1)/((-4)-1) = (-3)/(-5) = 3/5`

So

`lim_(xrarr-4)(x^2+5x+4)/(x^2+3x-4) = 3/5`