What is the limit of #(2x-1)/(4x^2-1)# as #x# approaches #-1/2#?

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Answer 1 · 2014-09-27T23:38:45

#lim_{x to -1/2}{2x-1}/{4x^2-1}# does not exist.

Let us evaluate the left-hand limit.

#lim_{x to -1/2"^-}{2x-1}/{4x^2-1}#

by factoring out the denominator,

#=lim_{x to -1/2"^-}{2x-1}/{(2x-1)(2x+1)}#

by cancelling out #(2x-1)#'s,

#=lim_{x to -1/2"^-}1/{2x+1}=1/{0^-} = -infty#

Let us evaluate the right-hand limit.

#lim_{x to -1/2"^+}{2x-1}/{4x^2-1}#

by factoring out the denominator,

#=lim_{x to -1/2"^+}{2x-1}/{(2x-1)(2x+1)}#

by cancelling out #(2x-1)#'s,

#=lim_{x to -1/2"^+}1/{2x+1}=1/{0^+} =+infty#

Hence, #lim_{x to -1/2}{2x-1}/{4x^2-1}# does not exist.