I solved it with L'Hopital's rule but I want to find out how can I solve it without using L'Hopital's rule.
Answer
limx→0x−tanxxtanx
=limx→0(xxtanx−tanxxtanx)
=limx→0(1tanx−1x)
=limx→0(cosxsinx−1x)
=limx→0(cosxxsinxx−1x)
=limx→0(cosxsinxx1x−1x)
=limx→0(1x−1x)
Because, limx→0cosx=1 and limx→0sinxx=1
=0
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