Bài giải:
d.
$lim_{x->1}\frac{x^2-3x+2}{x-1}$
=$lim_{x->1}\frac{(x-1)(x-2)}{x-1}$
=$lim_{x->1}x-2$
=$1-2$
=$-1$
e.
$lim_{x->-1}\frac{x^2+7x+6}{x^3+1}$
=$lim_{x->1}\frac{(x+1).(x+6)}{(x+1).(x^2-x+1)}$
=$lim_{x->1}\frac{x+6}{x^2-x+1}$
=$\frac{7}{1}$
=$7$