求极限问题，同仁帮帮忙！

（1）lim(x→1-), lnxln(1-x)
= lim(x→1-) [ln(1-x)]/[1/lnx] [这是0/0型未定式，应用洛必达法则]
= lim(x→1-) (1/x)/[-1/[(1-x).[ln(1-x)]^2]]
= lim(x→1-) [(x-1)[ln(1-x)]^2]/x [0.∞/1型，需做一定变换才能应用洛必达法则]
= lim(x→1-) [[ln(1-x)]^2/[x/(x-1)] [∞/∞型未定式，应用洛必达法则]
= lim(x→1-) [2ln(1-x)]/[1/(x-1)] [∞/∞型未定式，继续应用洛必达法则]
=2lim(x→1-) [1/(1-x)]/[-1/(x-1)^2]
=2lim(x→1-) (1-x)
=0

（3）lim(x→∞), (1+a/x)^x) [1^∞型]
=lim(x→∞) e^[x.ln(1+a/x)] [∞.0型，做一定变换]
=lim(x→∞) e^[ln(1+a/x)/[1/(1-x)]] [0/0型，用洛必达法则]
=e^[lim(x→∞) [a/(1+a/x)]]
=e^a

（4）lim(x→0+), x^ln(1+x)
=lim(x→0+) e^[lnx.ln(1+x)] [∞.0型，做一定变换]
=lim(x→0+) e^[ln(1+x)/(1/lnx)] [0/0型，用洛必达法则]
=e^[lim(x→0+) [1/(1+x)/[-(1/[x.(lnx)^2]]]
=e^[lim(x→0+) [-[x.(lnx)^2]/(1+x)]]
=e^[lim(x→0+) [(lnx)^2/[-(1+1/x)]]] [∞/∞型，用洛必达法则]
=e^[lim(x→0+) [2lnx.(1/x)/(1/x^2)]]
=e^[lim(x→0+) [2lnx/(1/x)]]
=e^[lim(x→0+) (-2x)
=1