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

Sep 27, 2014

${\lim}_{x \to - \frac{1}{2}} \frac{2 x - 1}{4 {x}^{2} - 1}$ 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 $\left(2 x - 1\right)$'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 $\left(2 x - 1\right)$'s,

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

Hence, ${\lim}_{x \to - \frac{1}{2}} \frac{2 x - 1}{4 {x}^{2} - 1}$ does not exist.