A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

watchmath

  • 5 years ago

Compute \[\lim_{x\to +\infty} x\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\]

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    even with the "x" out front?

  2. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, the problem is correct

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\lim_{x\to +\infty} x\times \frac{x}{x}\left(\sqrt{x^2+1}-\sqrt[3]{x^3+1}\right)\] \[ x^2 ( \sqrt{1 + \frac{1}{x^2}} + \sqrt[3]{1 + \frac{1}{x^3}})\] I think it must be \(\infty\)

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i meant with the x out front it seemed unlikely to be 0

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ishaan it is a minus sign. with a plus you would get infinity for sure

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it is minus

  7. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lost a minus sign mertsj, right\[x^2 ( \sqrt{1 + \frac{1}{x^2}} - \sqrt[3]{1 + \frac{1}{x^3}})=\infty(0)\]so it is still indeterminate

  8. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, ishaan, whoever it was lol

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh yeah, sorry. thanks satellite

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i will make a guess at a method, but it looks like a lot of work (algebra) i bet the usual gimmick will work of multiplying by the conjugate, but since you have powers of 1/2 and 1/3 you are going to have to use \[(a^6-b^6)=(a-b) (a+b) (a^2-a b+b^2) (a^2+a b+b^2)\]

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is just a guess though

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *Bookmark*

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    too much algebra for me, i can't do it

  14. Mertsj
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You can. You just don't want to.

  15. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think i might have it

  16. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hello myininaya. Still remember me? :)

  17. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i do i'm getting the limit is infinity is that right?

  18. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so i will show you what i did...

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @watchmath myininaya and i were just mentioning you the other day. sunday

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i said i missed your problems (not that i can do them) and no the answer is not infinity, but i cheated.

  21. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no, the limit is finite

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think the method i wrote would work if i could do the algebra

  23. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\lim_{x \rightarrow \infty}\frac{x((x^2+1)^\frac{1}{2}-(x^3+1))^\frac{1}{3}}{1} \cdot \frac{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\] \[\lim_{x \rightarrow \infty}\frac{x(x^2+1-(x^3+1)^\frac{2}{3})}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\] \[\lim_{x \rightarrow \infty}\frac{x^3+x-x(x^3+1)^\frac{2}{3}}{(x^2+1)^\frac{1}{2}+(x^3+1)^\frac{1}{3}}\] So if I divide both top and bottom by x I get \[\lim_{x \rightarrow \infty}\frac{x^2+1-(x^3+1)^\frac{2}{3}}{(\frac{x^2+1}{x^2})^\frac{1}{2}+(\frac{x^3+1}{x^3})^\frac{1}{3}}\] \[\lim_{x \rightarrow \infty}\frac{x^2+1-(x^3+1)^\frac{2}{3}}{(1+\frac{1}{x^2})^\frac{1}{2}+(1+\frac{1}{x^3})^\frac{1}{3}}\] so we have that the bottom goes to 2 as x goes to positive infinity but the top goes to infinity so that is why i conclude the answer was infinity Maybe I should play with the top some more and hey how are you watchman lol i have missed your problems

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @myininaya your numerator is wrong

  25. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    middle terms don't add to zero because the powers are not the same. i think it the same gimmick will work, multiply something to clear the radicals, but since one power is 1/2 and the other is 1/3 i am pretty sure you have to use \[(a^6-b^6)=(a-b) (a+b) (a^2-a b+b^2) (a^2+a b+b^2)\]to clear them

  26. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    satellite i disagree i think my numerator is right

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i am sorry, yes it is right, but you still can't get the limit that way

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here's a solution. Using the extension of the binomial theorem to arbitrary complex numbers, you have that \[(1+x^2)^\frac12=x+\frac12x^{-1}-\frac18x^{-3}+O(x^{-5})\] \[(1+x^3)^\frac13=x+\frac13x^{-2}+O(x^{-5})\] Thus, you have that \[\lim_{x\to \infty}x\left((1+x^2)^\frac12-(1+x^3)^\frac13\right)\] \[=\lim_{x\to\infty}x\left(\left(x+\frac12x^{-1}+O(x^{-3})\right)-\left(x+\frac13x^{-2}+O(x^{-5})\right)\right)\] \[=\lim_{x\to\infty}x\left(\frac12x^{-1}+O(x^{-2})\right)=\lim_{x\to\infty}\frac12+O(x^{-1})=\frac12\]

  29. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    imperialist is a showoff

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    pretty damned nice

  31. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    imperialist is correct. Any elementary solution?

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm pretty sure satellite's method is also a good way to approach it, but as myininaya can attest, the algebra is not very friendly to say the least. Good job though myininaya!

  33. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well i am still thinking of writing the difference of two sixth powers, the algebra is killing me

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i would say the above approach is much nicer, especially since i have filled up a piece of paper and probably have at least one mistake

  35. Mertsj
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why don't you brainiacs go help pre-algebra? He's been trying to get help most of the day.

  36. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you know numerically my top approaches 1

  37. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    algebraically my bottom approaches 2

  38. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if some how i can show without a numerical approach or graphical approach that the top approaches 1, then I would be dandy

  39. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where is the pre-algebra question? i can't find it

  40. Mertsj
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Updated 41 minutes ago.

  41. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So how can I show that \[\lim_{x \rightarrow \infty}(x^2+1-(x^3+1)^\frac{2}{3})=1?\]

  42. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm x^2 + 1 - (x^3+1)^(2/3) can't we ignore 1?

  43. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    couldn't i say \[x^3+1 \approx x^3\]

  44. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[(x^3)^\frac{2}{3}=x^2\]

  45. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x^2-x^2=0 we are left with 1

  46. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Mhm, sounds good to me!

  47. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think it sounds awesome! :)

  48. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If you can compute \[\lim_{x\to \infty} x\left(\sqrt[3]{x^3+1}-x\right)\] then you can compute the original problem :D.

  49. JamesJ
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (For what it's worth, I think Imperialist has it. I can't see a proof that doesn't rely on some version of a power series expansion. I've learnt the hard way never to say never, so perhaps there something else more elementary out there. Nonetheless, sometimes elementary problems have complicated solutions.)

  50. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    James what do you think about what I was saying?

  51. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We can say \[x^3 \approx x^3+1 \]

  52. JamesJ
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The trouble is how exactly is that notation defined? Intuitively, what you've done is right. But can you be sure that it makes sense and doesn't lead to contradictions. The advantage of what Imperialist has done is all the notation he's used is very defined and can be in such a way that we can be sure of our conclusions.

  53. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if you are expecting this problem to be answered elementary my suggestion is to split it into computing \(\lim x(\sqrt{x^2+1}-x)\) and \(\lim x(x-\sqrt{x^3+1})\)

  54. JamesJ
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cube root in that second expression, right?

  55. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, sorry for that

  56. JamesJ
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that's a nice idea. wish I'd thought of it.

  57. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.