求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限

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求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限

求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限

求lim(x→0+) ( 2/π*cosπ/2(1-x))/x的极限
lim(x→0+) { 2/π*cos[(π/2)(1-x)] } / x
= lim(x→0+) cos (π/2 - πx/2) / (πx/2)
= lim(x→0+) sin (πx/2) / (πx/2)
x→0+ 时 πx/2→0
则 lim(x→0+) sin (πx/2) / (πx/2)=1

0/0型,直接罗比达法则一次即可
lim(x→0+) ( 2/π*cosπ/2(1-x))/x
=lim(x→0+) sinπ/2(1-x)/1
=1