How do you evaluate the limit #(1-cosx)/tanx# as x approaches #0#?

1 Answer
Cesareo R.
Sep 28, 2016

#0#

Explanation:

Using de Moivre's identity

#e^(ix) = cos x+i sin x#

#(1-cosx)/tanx = (1-(e^(ix)+e^(-ix))/2)(e^(ix)+e^(-ix))/(e^(ix)-e^(-ix))=#
#((2-(e^(ix)+e^(-ix))))/(2(e^(ix)-e^(-ix)))(e^(ix)+e^(-ix))=#
#((e^(ix/2)-e^(-ix/2))^2(e^(ix)+e^(-ix)))/(2(e^(ix/2)+e^(-ix/2))(e^(ix/2)-e^(-ix/2))) = #
#=((e^(ix/2)-e^(-ix/2))(e^(ix)+e^(-ix)))/(2(e^(ix/2)+e^(-ix/2))) #

So

#lim_(x->0)(1-cosx)/tanx = lim_(x->0)((e^(ix/2)-e^(-ix/2))(e^(ix)+e^(-ix)))/(2(e^(ix/2)+e^(-ix/2)))=0#

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