Question

这个极限怎么算

Answer 1

设y=lim (sint/sinx)^(x/sint-sinx)
两边取对数
lny=lim (x/sint-sinx)*ln(sint/sinx)
=lim x*ln(sint/sinx)/(sint-sinx)
当t=x时，可知sint=sinx，sint/sinx=1
分子lnsint/sinx=ln1=0
分母sint-sinx=0
满足洛必达法则，注意这题中t是未知数
lny=lim x[cost/sinx*sinx/sint]/cost
=lim xcost/sint/cost
=lim x/sint
=x/sinx
y=e^(x/sinx)