Question
Evaluate:
$\lim\limits _{\text{x} \rightarrow \frac{1}{2}}\frac{4\text{x}^{2}-1}{2\text{x}-1}$

Answer

Given that $\lim\limits _{\text{x} \rightarrow \frac{1}{2}}\frac{4\text{x}^{2}-1}{2\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\frac{(2\text{x})^{2}-(1)^{2}}{2\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\frac{(2\text{x}+1)(2\text{x}-1)}{2\text{x}-1}$ 
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}(2\text{x}+1)$
Taking limit, We have
$=2 \times\frac{1}{2}+1$
$=1+1 $
$=2$
Hence, the answer is 2.

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