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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