Recent content by emilkh

Draw f(x) = x^2, then shift it down by 1, then draw f(x) = 1/(x^2 -1)

Show \sum_1^\infty\frac{x^n}{1+x^n} converges when x is in [0,1) \sum_1^\infty\frac{x^n}{1+x^n} = \sum_1^\infty\frac{1}{1+x^n} * x^n <= \sum_1^\infty\frac{1}{1} * x^n = \sum_1^\infty x^n The last sum is g-series, converges since r = x < 1

[edited for content], in my solutions it was 1, but somehow I copied formula wrong and the whole time assumes lim x^x = 0. I spend way too much time studying for finals... got to take break

Well great! This is where I am stuck. lim ln x * x was solved by switching it to lim ln x / (1/x) and taking derivatives, with lim ln x * x/x it's not going to work

lim ln x * (x^x) = lim ln x * lim x ^ x = (- infinity) * 0 = 0 as X -> 0+ So you want to tell me that lim (X^x^x) = lim ex^x *ln x = e^0 = 1? The limit suppose to be 0, .000001 ^ ( .000001 ^ .000001 ) = very very very small number (checked with calculator)

Could you elaborate more? I alredy tried this method and got stuck with lim (ln x) * (x^x), i could not solve it

Any ideas? lim XXX x-> 0+ I know how to do x^x: lim XX = lim ex * ln x = e0 = 0 lim x * ln x = lim ln x / (1/x) = lim (1/x) / (-1/x2) = lim -x = 0