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

1 Answer
George C.
Oct 6, 2016

#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#

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