Question

用洛必达法则求四个极限，在线等，谢谢

Answer 1

lim(x->1) (x^2-1)/(√x-1) (0/0)
=lim(x->1) 2x/[1/(2√x) ]
=lim(x->1) x^(3/2)
=1

lim(x->π) sinx/(x-π) (0/0)
=lim(x->π) cosx
=-1

lim(x->∞) 2^x/x^3 (∞/∞)
=lim(x->∞) (ln2)2^x/(3x^2) (∞/∞)
=lim(x->∞) (ln2)^2.2^x/(6x) (∞/∞)
=lim(x->∞) (ln2)^3.2^x/6
->∞

lim(x->0) (e^x -e^(-x) -2x)/(x-sinx) (0/0)
=lim(x->0) (e^x +e^(-x) -2)/(1-cosx) (0/0)
=lim(x->0) (e^x -e^(-x) )/sinx (0/0)
=lim(x->0) (e^x +e^(-x) )/cosx
=2