Question
$\lim _{x \rightarrow 1 / 2} \frac{4 x^2-1}{2 x-1}$ is equal to _________________

Answer

2 , because
$
\begin{aligned}
\lim _{x \rightarrow 1 / 2} \frac{4 x^2-1}{2 x-1} & =\lim _{x \rightarrow 1 / 2} \frac{(2 x)^2-(1)^2}{(2 x-1)} \\
& =\lim _{x \rightarrow 1 / 2} \frac{(2 x+1)(2 x-1)}{(2 x-1)} \\
& =\lim _{x \rightarrow 1 / 2}(2 x+1) \quad\left(x \neq \frac{1}{2}\right) \\
& =2 \times \frac{1}{2}+1=1+1=2 .
\end{aligned}
$

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