Evaluate: _ x 1 2 ( 8x-3 2x-1 - 4x^ 2 +1 4x^ 2 -1 )

Question

Evaluate:
$\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\Big(\frac{8\text{x}-3}{2\text{x}-1}-\frac{4\text{x}^{2}+1}{4\text{x}^{2}-1}\Big)$

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Answer

Given that $\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\Big(\frac{8\text{x}-3}{2\text{x}-1}-\frac{4\text{x}^{2}+1}{4\text{x}^{2}-1}\Big)$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2} }\Big[\frac{(8\text{x}-3)(2\text{x}-1)-(4\text{x}^{2}+1)}{(4\text{x}^{2}-1)}\Big]$
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\Big[\frac{16\text{x}^{2}-6\text{x}+8\text{x}-3-4\text{x}^{2}-1}{4\text{x}^{2}-1}\Big] $
$=\lim\limits_{\text{x} \rightarrow \frac{1}{2}}\frac{2(6\text{x}^{2}+\text{x}-2)}{4\text{x}^{2}-1}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{2[6\text{x}^{2}+4\text{x}-3\text{x}-2]}{(2\text{x}+1)(2\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{2[ 2\text{x}(3\text{x}+2)-1(3\text{x}+2)]}{(2\text{x}+1)(2\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{ 2(3\text{x}+2)(2\text{x}-1)]}{(2\text{x}+1)(2\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow\frac{1}{2}} \frac{ 2(3\text{x}+2)}{(2\text{x}+1)}$
Taking limit, we have
$=\frac{2\big(3\times\frac{1}{2}+2\big)}{2\times\frac{1}{2}+1}$
$=\frac{2\big(\frac{7}{2}\big)}{2}=\frac{7}{2}$
Hence, the required answer is $\frac{7}{2}.$

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