题目

求极限lim(1/x-cot x )
自己已经算出

提问时间：2020-12-22

答案

无论左极限,还是右极限,都是0.
lim [1/x - cotx]
x→0+
=lim (sinx - xcosx)/(xsinx) (0/0 型不定式)
x→0+
=lim (cosx - cosx + xsinx)/(sinx + xcosx) (使用了罗毕达法则)
x→0+
=lim (xsinx)/(sinx + xcosx)
x→0+
=lim 1/(1/x + cosx/sinx) (分子分母除以xsinx)
x→0+
= 1/(+∞ + ∞)
= 0
lim [1/x - cotx]
x→0-
=lim (sinx - xcosx)/(xsinx) (0/0 型不定式)
x→0-
=lim (cosx - cosx + xsinx)/(sinx + xcosx) (使用了罗毕达法则)
x→0-
=lim (xsinx)/(sinx + xcosx)
x→0-
=lim 1/(1/x + cosx/sinx) (分子分母除以xsinx)
x→0-
= 1/(-∞ - ∞)
= 0