Question

1 1grade please help​

Answer 1

Gawd that looks really hard

Answer 2

Answer:  7/2

Step-by-step explanation:

$\lim_{x \to \frac{1}{2}} \dfrac{8x-3}{2x-1}-\dfrac{4x^2+1}{4x^2-1}$

Multiply the left fraction by 2x+1 so both fractions have the same denominator.

$\lim_{x \to \frac{1}{2}} \dfrac{8x-3}{2x-1}\bigg(\dfrac{2x+1}{2x+1}\bigg)-\dfrac{4x^2+1}{4x^2-1}\\\\\\\lim_{x \to \frac{1}{2}} \dfrac{12x^2+2x-4}{4x^2-1}\\\\\\\lim_{x \to \frac{1}{2}} \dfrac{2(3x+2)(2x-1)}{(2x+1)(2x-1)}\\\\\\\lim_{x \to \frac{1}{2}} \dfrac{2(3x+2)}{2x+1}$

Directly input x = 1/2        $\dfrac{2[3(\frac{1}{2})+2]}{2(\frac{1}{2})+1}\quad =\dfrac{7}{2}$