Đáp án:
$\lim\limits_{x\to 1}\dfrac{x^{1992} + x - 2}{x^{1990} + x - 2}=\dfrac{1993}{1991}$
Giải thích các bước giải:
$A=\dfrac{x^{1992} + x - 2}{x^{1990} + x - 2}=\dfrac{x^{1992}-1 + x - 1}{x^{1990}-1 + x - 1}$
$=\dfrac{(x-1)(x^{1991} + x^{1990} +...+x+1)+(x-1)}{(x-1)(x^{1989}+x^{1987}+...+x+1) +( x - 1)}$
$=\dfrac{(x-1)(x^{1991}+x^{1990}+...+x+2)}{(x-1)(x^{1989}+x^{1987}+...+x+2)}$
$\lim\limits_{x\to 1}A=\dfrac{x^{1991}+x^{1990}+...+x+2}{x^{1989}+x^{1987}+...+x+2}$
$=\dfrac{1^{1991}+1^{1990}+...+1+2}{1^{1989}+1^{1987}+...+1+2}$
$=\dfrac{1993}{1991}$
