Question

Find $\underset{x\to \frac{1}{2}}{lim}\left(\frac{8x-3}{2x-1}-\frac{4x^2+1}{4x^2-1}\right)=$

• A. $\frac{7}{2}$
• B. $\frac{7}{3}$
• C. $\frac{5}{3}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{7}{2}$

The given function can be written in the form of

$\frac{(8x-3)(2x+1)-(4x^2+1)}{4x^2-1}$

Hence,

$\underset{x\to \frac{1}{2}}{lim}\frac{(8x-3)(2x+1)-(4x^2+1)}{4x^2-1}$

$=\underset{x\to \frac{1}{2}}{lim}\frac{12x^2+2x-4}{4x^2-1}$

Since, this is of the form $\frac{0}{0}$

We apply L-Hospital Rule to get,

$\underset{x\to \frac{1}{2}}{lim}\frac{24x+2}{8x}=\frac{14}{4}=\frac{7}{2}$