How to solve #lim_(xto0)(ln cotx)^tanx#?

1 Answer
1s2s2p
Mar 5, 2018

#lim_(x->0)(lncotx)^tanx=1#

Explanation:

#lim_(x->0)tanx=0#

#lim_(x->0^+)cotx=+oo#
#lim_(x->0^-)cotx=-oo#

#lim_(x->+oo)ln(x)=oo#

#oo^0=1# since #a^0=1,a!=0# (we'll say #a!=0#, since it get's a little bit complicated otherwise, some say it is 1, some say 0, others say it is undefined, etc.)

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