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

Answer 1

#0#

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))) #

#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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Answer 2

Isaiah Allen

To evaluate the limit (1-cosx)/tanx as x approaches 0, we can use the concept of trigonometric identities. By applying the identity sinx/tanx = cosx, we can rewrite the expression as sinx/(sinx/cosx), which simplifies to cosx. Therefore, the limit of (1-cosx)/tanx as x approaches 0 is equal to 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.