Question

Найдите предел


$\lim_{x \to \ 1/2}$ $\frac{8x^{3}-1 }{6x^{2}-5x+1 }$

Answer 1

$\displaystyle \lim_{x \to \frac{1}{2}}\frac{8x^3-1}{6x^2-5x+1}=\lim_{x \to \frac{1}{2}}\frac{(2x-1)(4x^2+2x+1)}{(2x-1)(3x-1)}=\lim_{x \to \frac{1}{2}}\frac{4x^2+2x+1}{3x-1}=\\ \\ \\ =\dfrac{1+1+1}{3\cdot \frac{1}{2}-1}=\frac{3\cdot2}{3-2}=6$