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What is #lim_(x->0) (1-cosx)^2/x # ?

1 Answer

Jun 8, 2017

#lim_"x->0" (1-cosx)^2/x = 0#

Explanation:

#lim_"x->0" (1-cosx)^2/x #

Here we have a limit which reduces to the indeterminate #0/0#, hence L'hopital's rule applies.

#:. lim_"x->0" (1-cosx)^2/x = lim_"x->0" (2*(1-cosx)*sinx)/1#

#=(2xx0xx0)/1 = 0#

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