M MechEng May 2010 129 0 WI - USA May 21, 2010 #1 Good afternoon all. I am getting back into calculs after a 5 year hiatus, and am finding that I seem to have forgotten some of the finer points. I was wondering if someone could walk me through the follwoing: lim, x->0, x^(sin(x)) Thanks much

Good afternoon all. I am getting back into calculs after a 5 year hiatus, and am finding that I seem to have forgotten some of the finer points. I was wondering if someone could walk me through the follwoing: lim, x->0, x^(sin(x)) Thanks much

Also sprach Zarathustra Dec 2009 1,506 434 Russia May 21, 2010 #2 + lim x^sinx = lim e^(sinx)lnx = e^lim((sinx)lnx =e^0=1 lim (x-->0+) sinx*lnx = 0, why is that? Try l'Hopital on ln(x)/(1/sin(x)). What does that give you? Remember lim sin(x)/x=1.

+ lim x^sinx = lim e^(sinx)lnx = e^lim((sinx)lnx =e^0=1 lim (x-->0+) sinx*lnx = 0, why is that? Try l'Hopital on ln(x)/(1/sin(x)). What does that give you? Remember lim sin(x)/x=1.

T Ted Feb 2010 240 64 China May 21, 2010 #3 It goes to 0^0 which is indeterminate. Put y=x^(sin x) --> ln(y) = (sin x) ln(x) Find the limit of ln(y) and let it be a. then the answer for your question is e^a.

It goes to 0^0 which is indeterminate. Put y=x^(sin x) --> ln(y) = (sin x) ln(x) Find the limit of ln(y) and let it be a. then the answer for your question is e^a.