Question

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

Answer 1

`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`