Question

1 1grade please help​

Answer 1

Gawd that looks really hard

Answer 2

Answer: 7/2

Explanation:

`\lim_{x \to (1)/(2)} (8x-3)/(2x-1)-(4x^2+1)/(4x^2-1)`

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

`\lim_{x \to (1)/(2)} (8x-3)/(2x-1)\bigg((2x+1)/(2x+1)\bigg)-(4x^2+1)/(4x^2-1)\\\\\\\lim_{x \to (1)/(2)} (12x^2+2x-4)/(4x^2-1)\\\\\\\lim_{x \to (1)/(2)} (2(3x+2)(2x-1))/((2x+1)(2x-1))\\\\\\\lim_{x \to (1)/(2)} (2(3x+2))/(2x+1)`

Directly input x = 1/2
`(2[3((1)/(2))+2])/(2((1)/(2))+1)\quad =(7)/(2)`