1.lim(x->1)[x^(1/(1-x))]=?
∵lim(x->1)[lnx/(1-x)]=lim(x->1)[(1/x)/(-1)] (0/0型極限,應(yīng)用羅比達(dá)法則)
=lim(x->1)[(-1)/x]
=-1
∴l(xiāng)im(x->1)[x^(1/(1-x))]=lim(x->1){e^[lnx/(1-x)]}
=e^{lim(x->1)[lnx/(1-x)]}
=e^(-1)
=1/e.
2.lim(x->0)[x^(tanx)]=?
∵lim(x->0)(tanx*lnx)=lim(x->0)[(sinx/x)(1/cosx)(lnx/(1/x))]
=[lim(x->0)(sinx/x)]*[lim(x->0)(1/cosx)]*[lim(x->0)(lnx/(1/x))]
=1*1*[lim(x->0)(lnx/(1/x))] (應(yīng)用重要極限)
=lim(x->0)(lnx/(1/x))
=lim(x->0)[(1/x/(-1/x²))] (∞/∞型極限,應(yīng)用羅比達(dá)法則)
=lim(x->0)(-x)
=0
∴l(xiāng)im(x->0)[x^(tanx)]=lim(x->0)[e^(tanx*lnx)]
=e^[lim(x->0)(tanx*lnx)]
=e^(0)
=1.