Question

Problema cu limite. Rezolvati va rog cat mai degraba ca am evaluare

Answer 1

 

$\displaystyle\\8x^3-1=(2x)^3-1=(2x-1)[(2x)^2+2x+1]\\6x^2-5x+1=6x^2-3x-2x+1 = 3x(2x-1)-(2x-1)=(2x-1)(3x-1)\\\\ \lim_{x \to \frac{1}{2}} \frac{2x^3-1}{6x^2-5x+1}=\\  \\=\lim_{x \to \frac{1}{2}} \frac{(2x-1)[(2x)^2+2x+1]}{(2x-1)(3x-1)}=~~\text{(Simplificam)}\\\\=\lim_{x \to \frac{1}{2}} \frac{(2x)^2+2x+1}{3x-1}=~~~\text{(Inlocuim pe x cu 1/2)}=$

$\displaystyle\\=\lim_{x \to \frac{1}{2}} \frac{(2\cdot \frac{1}{2} )^2+2\cdot \frac{1}{2} +1}{3\cdot \frac{1}{2} -1}=\frac{1+1+1}{\frac{3}{2}-1 }=\frac{3}{\frac{1}{2}}=\boxed{6}$