## anonymous 3 years ago More help with l'hospital's rule: can I evaluate a problem using l'hospital's rule when the following condition exists: lim(x-->infinity) (xe^(1/x)-x)

1. anonymous

Sorry, the condition, after I worked the problem is -1/infinity

2. hartnn

no, L'Hopital's can be applied only if you have the form 0/0 or $$\infty$$/$$\infty$$ so, what you can do is, (xe^(1/x)-x) =x( (e^(1/x)-1)) = (e^(1/x)-1) / (1/x) now its 0/0 form and you can apply L'Hopitals.

3. kirbykirby

you get a form infinity - infinity which is undefined

4. anonymous

Thank you, I set it up differently :(

5. anonymous

Also, aren't there other conditions for l'hospital's? For example 1^infinity?

6. hartnn

if you have 1^infinity form, you can apply log to both sides to get 0/0 form, but there are other ways to deal with 1^infinity form also.

7. hartnn

for L'Hopital's , only 0/0 or infinity/infinity should be there.

8. kirbykirby

You can only apply L'Hopital's rule when you have the form $0/0$ or $\infty/\infty$. When you have 1^infinity, you usually transform your by using ln (natural log) on both sides of your equation (set your limit as y = lim... blabla) then ln both sides

9. anonymous

I got infinity over infinity when I worked that problem, how did you get zero over zero?

10. hartnn

i meant 0/0 or infinity / infinity ...any one of the form you can get...

11. anonymous

I know, but after evaluating the above problem, I got -infinity/infinity, you said you got 0/0

12. anonymous

How did you work it differently?

13. hartnn

oh, 1/x =0, e^0 =1 1-1=0 <--numerator 1/x=0 <--denom.

14. anonymous

You are right! Thank you!!

15. anonymous

Thanks for the help. Keep your eyes peeled, I may need help differentiating it. Soon. I'll close this for now and open a new one if I need help differentiating it.

16. kirbykirby

(e^(1/x)-1) / (1/x)$\lim_{x \rightarrow \infty} \frac{e^{1/x}-1}{\frac{1}{x}}= \frac{\lim_{x \rightarrow \infty}e^{1/x}-1}{\lim_{x \rightarrow \infty}\frac{1}{x}}=\frac{e^{1/\infty}-1}{\frac{1}{\infty}}=\frac{e^0-1}{0}=0/0$

17. hartnn

ok, sure :)

18. anonymous

Thanks Kirby!

19. hartnn

one tip, before differentiating, substitute x =1/y so, that y-->0 then diff. num and denom. seperately. it'll much easier than to diff. by x.

20. anonymous

What? Why? I don't understand your tip at all... does infinity equal 1/0?

21. kirbykirby

It's the limit as x->0 of 1/x ... it's different than just "1/x" where x=0

22. hartnn

yes, it does.

23. kirbykirby

|dw:1358746300788:dw|

24. hartnn

x---> infinity x is very very large but not equal to infinity 1/x will be very very small, but not =0 -----> 1/x->0

25. anonymous

interesting

26. kirbykirby

As you go towards x->0 on the x-axis, you approach infinity because there is an asymptote at x=0. So, the graph is continuously stretching out infinitely along the asymptote

27. anonymous

It's hard for me to visualize. Why is that fraction easier to use than just x as it approaches infinity?

28. anonymous

So I have (-e^(1/x)/x^2)/-(1/x^2), which, using l'hospital's rule gives me 0/0. I differentiate that and I have

29. kirbykirby

x->0 from the right obviously, it approaches +infinity, but -infinity if it approaches 0 from the left

30. anonymous

1/infinity... which is zero, what do you think? Am I differentiating it correctly?

31. hartnn

see, if you differentiate directly, you'll need to differentiate, 1/x in the denominator, which will give you -1/x^2 , which again will be 0 for x->infinity. BUT, if you put y=1/x you'll have y in the denominator, and differentiating y will give u 1. got the point ?

32. anonymous

no... give me a second for my brain to process this

33. anonymous

Yeah, I don't see how putting zero in a denominator will ever give me one?

34. anonymous

if y is approaching zero, that is, and is in the denominator

35. hartnn

'putting zero in a denominator ' ??

36. anonymous

you said I should use y--->0 instead of 1/x as x---> infinity. If I plug y in my denominator, then I'll have zero.

37. anonymous

I'm supposing I actually have Zero idea what you're talking about... in the numerator

38. hartnn

wait, did you try differentiating without any substitution ? then you'll see what problems arise. then you put 1/x = y.

39. anonymous

I tried differentiating and I wound up with 1/infinity, which is zero. I'm wondering if I got it or if I botched it.

40. hartnn

umm....no, you don't get 1/infinity....show your steps....

41. anonymous

okay. I have to differentiate (-e^(1/x)/X^2)/(1/(x^2))

42. anonymous

I get -e^(1/x)/(x^2) * -x^2

43. hartnn

wait.

44. anonymous

that gives me -e^(1/x)/x^2 * -1

45. anonymous

I'm waiting

46. kirbykirby

(-e^(1/x)/X^2)/(1/(x^2))= -e^(1/x)

47. hartnn

numerator = e^(1/x) -1 ---> (-1/x^2) e^(1/x) denominator =1/x ------> (-1/x^2) overall = e^(1/x) yeah...

48. anonymous

so I do that before I differentiate

49. hartnn

you got (-1/x^2) e^(1/x) and (-1/x^2) after differentiating only...

50. hartnn

and then you can directly put 1/x = 0, e^0=1

51. anonymous

so I have 1/0, or indeterminate?

52. hartnn

you have 1 in the denominator now...not 0

53. anonymous

got it, so my answer is 1!

54. hartnn

yes!

55. anonymous

e^0/1?

56. anonymous

vunderbar! so I shouldn't have differentiated in the first place.

57. anonymous

I just evaluated it wrong. You are amazing. Who are you? Thanks a million!

58. hartnn

i am just another helper, glad i could help :) welcome ^_^