How do you find the limit of #(x^2-5x+4)/(x^2-2x-8)# as #x->4#?

1 Answer
Jim H
Oct 27, 2016

Factor, simplify and try again.

Explanation:

When we try to evaluate by substitution, we get

#lim_(xrarr4)(x^2-5x+4) = 0# and

#lim_(xrarr4)(x^2-2x-8) = 0# .

The initial form of #lim_(xrarr4)(x^2-5x+4) /(x^2-2x-8)# is the indeterminate form #0/0#.

Because #4# is a zero of both polynomials (the numerator and the denominator), we can be sure that #x-4# is a factor of both. Therefore we can simplify the ration and try again to evaluate the limit.

#lim_(xrarr4)(x^2-5x+4) /(x^2-2x-8) = lim_(xrarr4)((x-4)(x-1)) /((x-4)(x+2)) #

# = lim_(xrarr4)(x-1) /(x+2) = (4-1)/(4-2) = 3/2 #

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