[tex]\displaystyle\\\lim_{x\to2}\dfrac{x^3-2x^2+4x-8}{x^3+2x^2-4x-8}=\\\\\lim_{x\to2}\dfrac{\left(x^3-2x^2+4x-8\right)'}{\left(x^3+2x^2-4x-8\right)'}=\\\\\lim_{x\to2}\dfrac{3x^2-4x+4}{3x^2-4x-4}=\\\\\dfrac{3\cdot2^2-4\cdot2+4}{3\cdot2^2+4\cdot2-4}=\dfrac{12-8+4}{12+8-4}=\dfrac{8}{16}=\dfrac{1}{2}[/tex]
