Question

求第二题这两题的详系过程，要详细的过程！！

Answer 1

Answer 2

因为(x-1)/(x²+x-2)
=(x-1)/(x+2)(x-1)
=1/(x+2)所以
lim(x→1)(x-1)/(x²+x-2)
=lim(x→1)1/(x+2)
=1/(1+2)
=1/3
因为1/(x-2)-12/(x³-8)
=1/(x-2)-12/[(x-2)(x²+2x+4)]
=(x²+2x+4)/[(x-2)(x²+2x+4)]-12/[(x-2)(x²+2x+4)]
=(x²+2x+4-12)/[(x-2)(x²+2x+4)]
=(x²+2x-8)/[(x-2)(x²+2x+4)]
=(x-2)(x+4)/[(x-2)(x²+2x+4)]
=(x+4)/(x²+2x+4)所以
lim(x→2)1/(x-2)-12/(x³-8)
=lim(x→2)(x+4)/(x²+2x+4)
=(2+4)/(2²+2*2+4)
=6/(4+4+4)
=6/12
=1/2