Calc problem

Can someone show me how to solve this problem?  I'm lost.  Thanks alot for the help in advance!
lim(x->0+) of x^(1/ln(sinx))

looks like infinity to me. sin(x) is much smaller than x. 1/ln(sin(x)) gets bigger much faster than x gets smaller.

Ok, but how can I mathematically demonstrate that?

Looks like it is e, sir.
First you need to find lim(x->0+) of ln(x)/ln(sin(x)). If my reasoning was right, it should give you 1 (Involves using l'hopital's rule).
Second, as x^(1/ln(sin(x)) = e^(ln(x)/ln(sin(x))), use that the function e^x is continuous to conclude that the limit exists, and it is equal to e.

lim(x->0+) of ln(x)/ln(sin(x)). If my reasoning was right, it should give you 1 (Involves using l'hopital's rule).

no. that gives you -inf/0. l'hopital's rule doesn't apply. e^-inf does = infinity though. >>2 is correct

>>5

Pretty certain e^-inf = 0, didn't really read the rest of your post

>>5

sin(0) = 0, => lim(x->0+) ln(sin(x)) = lim(x->0+) ln(x) = -inf. You certainly failed basic trig or precalc, therefore, you shouldn't be reading this kind of threads.

>>6
>>7
holy fuck me in the ass. i don't know how i managed that :P