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

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#

#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#