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Evaluating the limit of a function

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm not sure how to type "lim x-> 0 out using this forum's math symbols...

    lim x→0 [[itex]\frac{1}{x\sqrt{1+x}}[/itex]-[itex]\frac{1}{x}[/itex]]


    2. Relevant equations



    3. The attempt at a solution

    Honestly, I'm just not seeing how to manipulate the equation to get it to a point where I can find the solution. This isn't graded and isn't even for any class I'm in...I'm just curious. It's been 10 years since I've done anything like this, so I'm trying to get up to speed before I take Calculus again in the fall.
     
  2. jcsd
  3. Feb 19, 2012 #2
    Since x is going to zero, I'd try a taylor series about x=0 ;)
     
  4. Feb 19, 2012 #3
    Haha, well, I'd love to. But since this is in one of the first few chapters in the book, long before Taylor series is introduced, there must be another way and I'd like do it that way first.
     
  5. Feb 19, 2012 #4
    oh.. :p

    In that case, try getting everything into one term, with a common denominator and see what happens
     
  6. Feb 19, 2012 #5
    I tried that and was just as lost. Maybe I did it wrong, but I got:

    lim x→0 [[itex]\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}[/itex]]

    If I did that correctly, I still don't know where to go from there.
     
  7. Feb 19, 2012 #6
    also try multiplying by 1 in a nice way :3
     
    Last edited: Feb 19, 2012
  8. Feb 19, 2012 #7
    Would you mind showing my the first step to take, so I can go on from there? I would appreciate it.
     
  9. Feb 19, 2012 #8

    Dick

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    Now try multiplying numerator and denominator by [itex]1+\sqrt{1+x}[/itex]. Expand out the numerator.
     
  10. Feb 19, 2012 #9
    No, the expansion was done correctly.

    Try rationalizing the numerator by multiplying by ##\frac{1 + \sqrt{1 + x}}{1 + \sqrt{1 + x}}##

    Also, Taylor series are used for approximating functions, not evaluating limits.

    Edit: ninjas on this board!
     
  11. Feb 19, 2012 #10
    Wow, so that was the trick. I carried out the math from there and caclulator/book confirms the answer. Thank you all so much. That was far simpler than it seemed.
     
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