What is the limit of #[arctan(e^b) - pi/4]# as b approaches infinity?

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Answer 1 · 2017-03-03T04:12:57

#lim_(brarroo)[arctan(e^b)-pi/4]=pi/2-pi/4=pi/4#

Note that as #x# approaches infinity, the #arctan(x)# function approaches #pi/2#.

We can reason through this by saying that:

#lim_(xrarrpi/2)tan(x)=oo#

So

#lim_(xrarroo)arctan(x)=pi/2#

So, we have

#lim_(brarroo)[arctan(e^b)-pi/4]=pi/2-pi/4=pi/4#