## anonymous 5 years ago anyone good at limits? lim(1+2/x)^x x-(infinity)

1. anonymous

$e^{-2}$is my guess

2. anonymous

any work to go with it?

3. myininaya

he might be right i remember that limit i think it was either as x->0 or infinity cant remember

4. anonymous

lim(1+3h)^(1/h) h->0

5. anonymous

sorry it is e^2

6. anonymous

$e=lim_{x->\infty}(1+\frac{1}{x})^x$

7. myininaya

if he is right about that one then this second one should be e^(1/[3h]) i believe

8. anonymous

and so of course $e^2=lim_{x->\infty}(1+\frac{2}{x})^x$

9. anonymous

you other one is e^3

10. anonymous

for exactly the same reason

11. anonymous

so then lim 3^(-x) x->+(infinity)

12. anonymous

if you want we can work this out step by step.

13. anonymous

sure,

14. anonymous

step one is take the log get $ln(1+\frac{1}{x})^x=xln(1+\frac{1}{x})$

15. myininaya

let 1/u=3h as h->0 then u->infinity so we have as u->infinity then (1+1/u)^u->e^u but u=1/(3h) so the limit is e^{1/(3h)}

16. myininaya

oh wiat nvm

17. anonymous

wait. myininaya i think it is just e^3

18. anonymous

we do it the donkey way

19. anonymous

take the log, take the limit using l'hopital, see that we get 3 and then conclude that it is e^3

20. anonymous

which one do you want to do?

21. myininaya

u r right

22. myininaya

its not fair u shouldn't always win

23. anonymous

fancy which one would you like worked out?

24. anonymous

lim 3^-x x-> +(infinty

25. myininaya

i want to see both :)

26. anonymous

$lim_{x->\infty}3^{-x}$

27. anonymous

nothing to that one. that is $lim_{x->\infty}\frac{1}{3^x}=0$

28. anonymous

this is the pain: (1+2/x)^x

29. anonymous

but we knock it out

30. anonymous

why 0

31. anonymous

as x gets large 3^x gets really large. huge denominator,1 in the numerator, very close to zero

32. anonymous

back to (1+2/x)^x because i gotta go

33. anonymous

limx-.2+ ln(x-2)

34. anonymous

take the log get xln(1+2/x) take the limit as x -> infinity get infinity times 0 so rewrite

35. anonymous

as $\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}$

36. anonymous

now we have 0/0 so use l'hopital

37. anonymous

ooo ojkak

38. anonymous

ooo ojkak

39. anonymous

i made a mistake and wrote 1 instead of 2. when you take the derivative of to bottom you get $-\frac{1}{x^2}$

40. anonymous

when you take the derivative of the top you get $\frac{1}{1+\frac{2}{x}}\times-2\frac{1}{x^2}$

41. anonymous

everything cancels leaving you with $\frac{2}{1+\frac{1}{x^2}}$

42. anonymous

then let x -> infinity and get 2

43. anonymous

and since we took the log in the first step the answer is not 2, but e^2

44. anonymous

so many questions, so little time