L'Hopital's Rule limits.

[akaich0u]

We're currently studying L'Hopital's Rule for limits in Calculus BC, particularly in application to logs. I just have a few questions:

Comments

Re: Logrithmic limits

[akaich0u]

Yes, that helped :) Thank you!

[korean_guy_01]

#23 is tricky, but it's a pretty common L'Hopital problem

x ln(x)
= [ln(x)]/(1/x)

f'/g' = (1/x)/(-1/x2)
Flip the denominator, becomes the numerator, cancel an x on top and bottom
= -x

Apply the Limit and you get 0

#33

Let L = the value of the Limit

ln(L) = (1/x)*ln(ln(x)) = [ln(ln(x))]/x

f'/g' = {[ln(x)]/x)}/1
= {[ln(x)]/x)}

r'/s' = (1/x)/1
= 1/x

Apply the Limit and you get 0. However, this equals ln(L), and L is your true answer. ln(L) = 0 -> L = 1 (since e^0 = 1)

[akaich0u]

Thanks! I think what got me really confused in 23 was the changing it to a quotient to apply L'Hopital's Rule..I totally just forgot the basic arithmetic of dividing by the reciprocal of what you multiply by XD 33 is also clear now; for limits with powers, you plug the limit back in as a power for e. I appreciate your help! :)