Giải thích các bước giải:
$e.\lim_{x\to 1}\dfrac{x^3-3x+2}{x^4-4x+3}$
$=\lim_{x\to 1}\dfrac{(x-1)^2(x+2)}{(x-1)^2(x^2+2x+3)}$
$=\lim_{x\to 1}\dfrac{x+2}{x^2+2x+3}$
$=\dfrac{1+2}{1+2+3}=\dfrac 12$
$k.\lim_{x\to 1}\dfrac{2x^4+8x^3+7x^2-4x-4}{3x^3+14x^2+20x+8}$
$=\lim_{x\to 1}\dfrac{(x+2)^2(2x^2-1)}{(x+2)^2(3x+2)}$
$=\lim_{x\to 1}\dfrac{2x^2-1}{3x+2}$
$=\dfrac{2-1}{3+2}=\dfrac 15$
