We have to find the value of
$\lim _{x \rightarrow \frac{1}{2}} \frac{8 x^3-1}{16 x^4-1}$
When $x =\frac{1}{2}$, the expression $\frac{8 x^3-1}{16 x^4-1}$ assumes
the for $n \frac{0}{0}$.
Therefore, $\left(x-\frac{1}{2}\right)$ or, $2 x -1$ is a factor common to
numerator and denominator.
Factorizing the numerator and denominator, we obtain;
$\lim _{x \rightarrow \frac{1}{2}} \frac{8 x^3-1}{16 x^4-1}\left(\frac{0}{0} \text { form }\right)$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x)^3-1^3}{\left(4 x^2\right)^2-1^2}$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)\left(4 x^2+2 x+1\right)}{\left(4 x^2+1\right)\left(4 x^2-1\right)}\left(\frac{0}{0} \text { form }\right)$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)\left(4 x^2+2 x+1\right)}{\left(4 x^2+1\right)(2 x-1)(2 x+1)}$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^2+2 x+1}{\left(4 x^2+1\right)(2 x+1)}=\frac{3}{4}$
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