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