Question

Answer 1

A=\(lim\left(x->2\right)\dfrac{2x^2-5x+2}{x^3-8}=lim\left(x->2\right)\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=lim\left(x->2\right)\dfrac{2x-1}{x^2+2x+4}\\ =\dfrac{2\cdot2-1}{2^2+2\cdot2+4}=\dfrac{3}{12}=\dfrac{1}{4}\)