## anonymous 5 years ago Lim(x->0)[1/x]^sinx

1. anonymous

that form is $\infty$^0, so u have to use exp function

2. anonymous

3. anonymous

Why don't you do a series expansion in x and then send the function to zero?

4. anonymous

Actually, I should have worked it out before opening my mouth.

5. anonymous

exp[lim(x->0) sin x . 1/x] exp[lim(x->0) (ln x)/sin x] and then do l'hospital

6. anonymous

Actually, what I said will work, but you'd have to prove other limits first.

7. anonymous

i do thinking about the concept before typing

8. anonymous

What are you thinking, iamignorant?

9. anonymous

Yes, I have got the idea that suzi gave. With that idea I am coming at e^(0x1)

10. anonymous

So that is the answer that I needed. Actually I did everything he did except applying lHospital

11. anonymous

ok

12. anonymous

Thank you all

13. anonymous

np

14. anonymous

is there a rank higher than HERO? =D like a grandmaster or something

15. anonymous

Who knows? This site's operations are a mystery half the time.

16. anonymous

i think GOD

17. anonymous

=D LOL

18. anonymous

19. anonymous

Incidentally, I set the RHS equal to y then took the log of both sides, put the result into an indeterminate form, used L'Hopital's rule and took the limit to zero after I suggested expanding the function. Just sayin'...

20. anonymous

What's going on BMF?