How do you find the limit #lim_(x->3^+)|3-x|/(x^2-2x-3)# ?

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Answer 1 · 2014-10-05T13:19:07

#lim_{x to 3^+}{|3-x|}/{x^2-2x-3}#

Note that since #x# approaches #3# from the right, #(3-x)# is negative, which means that #|3-x|=-(3-x)=x-3#.

by removing the absolute value sign and factoring out the denominator,

#=lim_{x to 3^+}{x-3}/{(x-3)(x+1)}#

by cancelling out #(x-3)#'s,

#=lim_{x to 3^+}1/{x+1}=1/{3+1}=1/4#

I hope that this was helpful.