lim x→1 (1-x)tan[(πx)/2]用洛必达法则如何求解我是这样解的:原式=(1-x)/1/tan(πx/2) =-1/1/(π/2)*[sec(πx/2)]^2 =-(π/2)/[cos(πx/2)]^2 =无穷大?答案是2/π 请问正确做法是什么?
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lim x→1 (1-x)tan[(πx)/2]用洛必达法则如何求解我是这样解的:原式=(1-x)/1/tan(πx/2) =-1/1/(π/2)*[sec(πx/2)]^2 =-(π/2)/[cos(πx/2)]^2 =无穷大?答案是2/π 请问正确做法是什么?
lim x→1 (1-x)tan[(πx)/2]用洛必达法则如何求解
我是这样解的:
原式=(1-x)/1/tan(πx/2)
=-1/1/(π/2)*[sec(πx/2)]^2
=-(π/2)/[cos(πx/2)]^2
=无穷大?
答案是2/π
请问正确做法是什么?
lim x→1 (1-x)tan[(πx)/2]用洛必达法则如何求解我是这样解的:原式=(1-x)/1/tan(πx/2) =-1/1/(π/2)*[sec(πx/2)]^2 =-(π/2)/[cos(πx/2)]^2 =无穷大?答案是2/π 请问正确做法是什么?
原极限
=lim(1-x)/cot(πx/2)
=lim(-1)/[-(π/2)(csc(πx/2))^2]
=lim[2/π(csc(πx/2))^2]
=2/π
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