Question

求大神做下这道题目 步骤详细 清楚

Answer 1

2、
lim(x→1)f(x)
=lim(x→1)lncos(x-1)/[1-sin(πx/2)]
=lim(x→1)[-sin(x-1)/cos(x-1)]/[-π/2cos(πx/2)]
=lim(x→1)(2/π)sin(x-1)/[cos(x-1)cos(πx/2)]
=lim(x→1)(2/π)cos(x-1)/[-sin(x-1)cos(πx/2)+cos(x-1)(-π/2)sin(πx/2)]
=(2/π)cos(1-1)/[-sin(1-1)cos(π/2)-π/2cos(1-1)sin(π/2)]
=(2/π)/(-π/2)
=-4/π^2
≠f(1)
f(x)在x=1处不连续。
注：^2——表示平方。