Question

第九，十题怎么做，在线等急

Answer 1

(9)
lim (1-x)tan(πx/2)
x→1
=lim (1-x)/cot(πx/2)
x→1
=lim (1-x)'/[cot(πx/2)]'
x→1
=lim (-1)/[-csc²(πx/2)·(π/2)]
x→1
=lim (2/π)sin²(πx/2)
x→1
=(2/π)·sin²(π/2)
=(2/π)·1
=2/π
(10)
lim (tanx-sinx)/x³
x→0
=lim (tanx-sinx)'/(x³)'
x→0
=lim (sec²x-cosx)'/(3x²)'
x→0
=lim (2sec²xtanx+sinx)'/(3·2x)'
x→0
=lim (4tan²xsec²x+2sec⁴x)/6
x→0
=lim (2tan²xsec²x+sec⁴x)/3
x→0
=(2·0·1+1)/3
=⅓

两题都是运用洛必达法则的典型题目。

Answer 2