Question

limit of x^(sinx) as x approaches 0 from the right. how can i rewrite sin(x)ln(x) so that I get the indeterminate form of 0/0 which will allow me to apply L'hospital's rule? I tried lnx/(1/sinx) but im not getting it because ln(0) is like undefined.

Answer 1

ok you got
\[\sin(x)\ln(x)\] as a first step, now it is in the form
\[0\times \infty\]

Answer 2

usual gimmick is to rewrite with algebra as
\[\frac{\ln(x)}{\frac{1}{\sin(x)}}\] or if you prefer
\[\frac{\ln(x)}{\csc(x)}\] and now it is in the form \[\frac{\infty}{\infty}\]

Answer 3

hmm that's what i had but I didn't know that it was in infinity/infinity form. thank you.

Answer 4

here is a useful trick

Answer 5

sin(x) ~ x , for x near zero

Answer 6

therefore

limit x^(sinx) -> limit x ^x , as x-> 0+

Answer 7

and limit x^x as ->0+ is 1