A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

unimatix

  • one year ago

Compute the standard parts: (sqrt(H+1))/(sqrt(2H)+sqrt(H-1)), H positive infinite

  • This Question is Closed
  1. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\large\color{blue}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x-1}}}\) can I interpret your question like this?

  2. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I believe that would be the same. Kind of trying to teach myself. The answer is given as 1 / ( 1 + sqrt(2) ) OR sqrt(2) - 1. I don't know how to solve it though.

  3. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    rationalize the denominator

  4. welshfella
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  5. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441555515323:dw|

  6. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441555908335:dw|

  7. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry this looks a bit messy. all three responses are the same . Rationalize the denominator. can you simplify the above drawing?

  8. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's fine. I'm trying to rationalize the denominator. You can simplify the denominator to X-1

  9. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x + 1

  10. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah it would be x +1.

  11. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    In the numerator I have sqrt( 2x^2+2x ) - sqrt (x^2-1)

  12. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441556580501:dw|

  13. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    needs more

  14. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess you could combine those to a single sqrt?

  15. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441556731148:dw|

  16. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    trying to work on the top

  17. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not really sure how you got to what is at the bottom in black

  18. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    work in progress however (x-1)^2 = x^2 -2 x + 1 thinking about difference of 2 squares

  19. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't really understand how in the third step you swapped the place sqrt(2x) with the sqrt(x+1). Wouldn't that give you a numerator not equal to the one before?

  20. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nevermind that haha

  21. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @SolomonZelman that was very impressive can you simplify my numerator think difference of 2 squares then divide by (x+1)

  22. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Just divide top and bottom by sqrt(2H)

  23. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\frac{\frac{H}{2H}+\frac{1}{2H}}{\sqrt{\frac{2H}{H}}+\sqrt{\frac{H}{2H}-\frac{1}{H}}}\]

  24. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    look at each term

  25. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oops forgot to right a square root thingy on top

  26. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    but you should get the idea

  27. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441558647746:dw|

  28. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @freckles can you please simplify this |dw:1441559138774:dw|

  29. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @triciaal I've been following what you are doing and am stuck at the same place.

  30. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\frac{\sqrt{H+1}}{\sqrt{2H}+\sqrt{H-1}} \cdot \frac{\frac{1}{\sqrt{2H}}}{\frac{1}{\sqrt{2H}}} \\ =\frac{\frac{\sqrt{H+1}}{\sqrt{2H}}}{\frac{\sqrt{2H}+\sqrt{H-1}}{\sqrt{2H}}} \\ =\frac{\sqrt{\frac{H}{2H}+\frac{1}{2H}}}{\frac{\sqrt{2H}}{\sqrt{2H}}+\sqrt{\frac{H}{2H}-\frac{1}{2H}}}\]

  31. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now just take the limit of each term as H goes to infinity

  32. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441559208589:dw|

  33. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    since everybody seems really confused on to do this... we have H tends to infinity \[\frac{\sqrt{\frac{H}{2H}+\frac{1}{2H}}}{\frac{\sqrt{2H}}{\sqrt{2H}}+\sqrt{\frac{H}{2H}-\frac{1}{2H}}}=\frac{\sqrt{\frac{1}{2}+0}}{1+\sqrt{\frac{1}{2}-0}}\] you can play with this answer to make it prettier

  34. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441559573245:dw|

  35. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    in the limit set equal square each side cancel common factor

  36. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you just have to divide top and bottom sqrt(2H) that is the key here

  37. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the hint was that H tends to infinity

  38. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @unimatix do you understand what I did above?

  39. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ freckles Your answer makes a lot of sense. What you made you decide to multiply by 1/sqrt(2H) though?

  40. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this is similar to having fractions where both numerator and denominator are polynomials when you have a polynomial over a polynomial you divide by the term on bottom that has highest exponent for the numerator and denominator that is of course when you are doing the whole x tends to positive or negative infinity this question is very similar except you do not have polynomials on top and bottom in a sense sqrt(H) was the term we are talking about I chose to divide by sqrt(2H) but sqrt(H) would work just as well or faster

  41. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and what I mean by faster is that you will see your answer more clearly because in the previous answer I gave you have to multiply top nad bottom by sqrt(2) to see you have the same answer

  42. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441560259041:dw|

  43. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\frac{\sqrt{\frac{1}{2}}}{1+\sqrt{\frac{1}{2}}} =\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \\=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ =\frac{\frac{\sqrt{2}}{\sqrt{2}}}{\sqrt{2}+\frac{\sqrt{2}}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1}\]

  44. triciaal
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1441560392712:dw|

  45. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\lim_{x \rightarrow \infty} \frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x-1}} \text{ instead we could have divided \top and bottom by } \sqrt{x} \\ \lim_{x \rightarrow \infty}\frac{\frac{\sqrt{x+1}}{\sqrt{x}}}{\frac{\sqrt{2x}}{\sqrt{x}}+\frac{\sqrt{x-1}}{\sqrt{x}}}=\lim_{x \rightarrow \infty} \frac{\sqrt{\frac{x}{x}+\frac{1}{x}}}{\sqrt{\frac{2x}{x}}+\sqrt{\frac{x}{x}-\frac{1}{x}}}=\frac{\sqrt{1+0}}{\sqrt{2}+\sqrt{1-0}}\]

  46. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks freckles, I was really getting frustrated! And thanks to triciaal and Solomon as well!

  47. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    similar problem: \[\lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \\ \lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\ \lim_{x \rightarrow \infty}\frac{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}{\frac{3x^2}{x^2}+\frac{1}{x^2}}=\frac{2+0+0}{3+0}=\frac{2}{3}\]

  48. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    WAHOO! I feel like I understand it now. Just have to remember to divide by polynomial with the highest power in the denominator.

  49. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    another similar problem: \[\lim_{x \rightarrow \infty}\frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \\ \\ \\\ \lim_{x \rightarrow \infty} \frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}{\sqrt{\frac{3x^2}{x^2}+\frac{1}{x^2}}} \\ =\frac{\sqrt{2+0+0}}{\sqrt{3+0}}=\frac{\sqrt{2}}{\sqrt{3}}\]

  50. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was able to work through it on my own and got the same answer!

  51. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\lim_{x \rightarrow \infty} \frac{\sqrt{2x+1}-1}{x} \\ \text{ here we have } x>0 \text{ so } \sqrt{x^2}=x \\ \text{ divide \top and bottom by } x \\ \text{ and we will come back \in use this one fact } \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{x}-\frac{1}{x}}{\frac{x}{x}} \\ \text{ now for the } \sqrt{x^2}=x \\ \text{ we can rewrite that one } x \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{\sqrt{x^2}}-\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x+1}{x^2}}-\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x}{x^2}-\frac{1}{x^2}}-\frac{1}{x}}{\frac{x}{x}} =\frac{\sqrt{0-0}-0}{1}=\frac{0}{1}=0\]

  52. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oops forgot to label that another example

  53. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and also fantastic

  54. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    these can get tricky when dividing top and bottom by sqrt(x^2) because sqrt(x^2)=x when x>0 and sqrt(x^2)=-x when x<0 just something to remember when you come face to face with it I'm sure you will

  55. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    some awesome notes to read: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

  56. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    example 3 is kind of example of what I'm talking about

  57. unimatix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright I'll read over it. Thanks so much!

  58. 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.