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

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Answer 1 · 2017-06-08T01:44:18

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

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

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

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

$=(2xx0xx0)/1 = 0$

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