Đáp án:
a) $\displaystyle\lim_{x \to -1} \dfrac{x^5+1}{x^3+1} = \dfrac{5}{3}$
b) $\displaystyle\lim_{x \to 1} \dfrac{x^3-1}{x^2-1} = \dfrac{3}{2}$
Giải thích các bước giải:
a) Ta có
$\displaystyle\lim_{x \to -1} \dfrac{x^5+1}{x^3+1} = \displaystyle\lim_{x \to -1} \dfrac{(x+1)(x^4 - x^3 + x^2 - x + 1)}{(x+1)(x^2 - x + 1)}$
$= \displaystyle\lim_{x \to -1} \dfrac{x^4 - x^3 + x^2 - x + 1}{x^2 - x + 1}$
$= \dfrac{1 +1 + 1 + 1 + 1}{1 + 1 + 1} = \dfrac{5}{3}$
b) Ta có
$\displaystyle\lim_{x \to 1} \dfrac{x^3-1}{x^2-1} = \displaystyle\lim_{x \to 1} \dfrac{(x-1)(x^2 +x + 1)}{(x-1)(x+1)}$
$= \displaystyle\lim_{x \to 1} \dfrac{x^2 + x + 1}{x+1}$
$= \dfrac{1 + 1 + 1}{1 + 1} = \dfrac{3}{2}$
