Question #5a865

1 Answer
Andrea S.
Feb 27, 2018

#lim_(x->0) floor(cosx) = 0#

Explanation:

In fact for any #x in (0,delta)# with #delta < pi/2# we have:

#0 < cosx < 1#

and then:

#floor(cosx) = 0#

so:

#lim_(x->0^+) floor(cosx) = 0#

and similarly:

#lim_(x->0^-) floor(cosx) = 0#

which imply:

#lim_(x->0) floor(cosx) = 0#

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