Expert: Sherman D. - 11/16/2005

Question
use l'hopital's rule to find the limits.

1) lim x-->0 (1/x^2)^x
2) lim x-->infinity (ln 2x - ln (x+1))

Answer
Sorry it took so long, but first i needed to understand how to work it, so i had to ask someone else to help me understand it.

1.)
(1/x^2)^x =
x^(-2x) =
(-2lnx)/(1/x) =
(-2'lnx + (-2lnx')/(1/(x^2)) =
(0lnx - 2lnx')/(1/(x^2)) =
(0lnx - (2/x))/(1/(x^2)) =
(-2/x)/(1/(x^2)) =
(-2x^2)/x =
-2x =
e^(-2x)

Limit is 0, since anything raised by a value can never equal 0.

-----------------------

2.)
ln(2x) - ln(x + 1) =
ln((2x)/(x + 1)) =
ln((2x)'/(x + 1)') =
ln((2'x + 2x')/1) =
ln((0x + 2)/1) =
ln2

limit is ln(2)