Find limit as x -> 0 of (1-cos(x))^2x using L'Hospital's Rule

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Find limit as x -> 0 of (1-cos(x))^2x using L'Hospital's Rule

Mathematics
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uh have you typed the problem correctly? I don't see division anywhere
There is not division yet, you have to take ln first
Oh the whole thing is raised to 2x ok

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You ln both sides as you said. so let y=(1-cosx)^2x. so lny=2xln(1-cosx) so now you are solving the limit of lny as x approaches 0. You have 0*infinity, so you change one into a quotient to get into a fraction. So 2ln(1-cosx)/(1/x). Now apply l'hopital multiple times and keep on simplifying until you no longer get an indeterminate form. You eventually find out the limit equals 0. So the limit for lny=0. Now recall, e^lny=y. So lim for y= e^(lim for lny) So solving the limit for lny gives us the exponent in the e. So the limit equals e^0=1.

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