What is the limit of $\frac{2 x - 1}{4 x^{2} - 1}$ $(2x-1)/(4x^2-1)$ as $x$ $x$ approaches $- \frac{1}{2}$ $-1/2$ ?

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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.

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