Question

用罗必达法则求下列极限

需详细步骤

Answer 1

1、原式=lim(x->1) 1/mx^(m-1)

=1/m
2、原式=lim(x->π/2+0) [1/(x-π/2)]/sec^2x
=lim(x->π/2+0) cos^2x/(x-π/2)
=lim(x->π/2+0) -2cosxsinx
=0
3、原式=lim(x->0) ln(1+x^2)*cosx/(1-cos^2x)
=lim(x->0) ln(1+x^2)/sin^2x
=lim(x->0) [2x/(1+x^2)]/2sinxcosx
=lim(x->0) x/(1+x^2)sinx
=lim(x->0) 1/[2xsinx+(1+x^2)cosx]
=1
4、原式=0*1=0
5、令t=1/x
原式=lim(t->0) (a^t-1)/t
=lim(t->0) lna*a^t
=lna
6、原式=lim(x->0+0) e^[tanx*ln(1/x)]
=e^lim(x->0+0) (-lnx/cotx)
=e^lim(x->0+0) (1/x)/csc^2x
=e^lim(x->0+0) (sin^2x)/x
=e^lim(x->0+0) 2sinxcosx
=e^0
=1