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Homework Help: Proof of limit involving square root

  1. Nov 11, 2004 #1
    Hello all

    I am having trouble proving the limit of the following:

    lim sqrt(( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) = 1/2
    n --> 00

    I tried using the fact the the limit of the first factor as n approaches infinity is 0. Then I tried expressing the first factor as

    1 / sqrt(n+1) + sqrt(n) and doing the same thing for the other

    factor. However I always get stuck.

    Any help would be greatly appreciated!
  2. jcsd
  3. Nov 11, 2004 #2


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    As you have written the expression the limit does not exist. I suspect you meant something else.
  4. Nov 12, 2004 #3
    lim (sqrt( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) = 1/2
    n --> 00
  5. Nov 12, 2004 #4
    I trhink so:
    lim (sqrt( n+1) - sqrt(n)) * sqrt(n+ 1/2 ) =
    =lim (sqrt( n+1) - sqrt(n)) *(sqrt( n+1) + sqrt(n)) * sqrt(n+ 1/2 ) /(sqrt( n+1) + sqrt(n)) = lim sqrt(n+ 1/2 )/(sqrt( n+1) + sqrt(n))=1/2
  6. Nov 12, 2004 #5


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    [tex]\lim_{n\rightarrow \infty} (\sqrt{n+1} - \sqrt{n}) \sqrt{n+\frac{1}{2}}=[/tex]
    [tex]\lim_{n\rightarrow \infty} \frac{((n+1)-n)\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=[/tex]
    [tex]\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}}=[/tex]

    [tex]2 \sqrt{n+1} > \sqrt{n+1} + \sqrt{n} > 2 \sqrt{n}[/tex]
    [tex]\frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n+1}} < \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} <
    [tex]\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n+1}} \leq \lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq \frac{1}{2}\lim_{n\rightarrow \infty}\frac{\sqrt{n+\frac{1}{2}}}{2\sqrt{n}} [/tex]
    [tex]\lim_{n\rightarrow \infty} \frac{1}{2}\sqrt{\frac{n+\frac{1}{2}}{n+1}} \leq \lim_{n\rightarrow \infty}\frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq \frac{1}{2}\lim_{n\rightarrow \infty}\sqrt{\frac{n+\frac{1}{2}}{n}}[/tex]
    [tex]\frac{1}{2}\lim_{n\rightarrow \infty} \sqrt{1 - \frac{\frac{1}{2}}{n+1}} \leq \lim_{n\rightarrow \infty} \frac{1}{2} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq\lim_{n\rightarrow \infty} \frac{1}{2}\sqrt{1 + \frac{\frac{1}{2}}{n}}[/tex]
    But now the limits on the RHS and LHS are pretty obviously 1 so we have:
    [tex]\frac{1}{2} \leq \lim_{n\rightarrow \infty}\frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} \leq \frac{1}{2}[/tex]
    so the limit is [tex]\frac{1}{2}[/tex]
  7. Nov 13, 2004 #6
    Thanks a lot for the very elegant solution!!!
  8. Nov 14, 2004 #7
    Actually I'd just stop there (I'm not saying Nate's solution is large or anything but here's another way to "see" where the limit is going). I'd then divide the numerator and the denominator by the square root of n to get

    [tex]\lim_{n\rightarrow \infty} \frac{\sqrt{n+\frac{1}{2}}}{\sqrt{n+1}+\sqrt{n}} = \lim_{n\rightarrow \infty} \frac{\sqrt{1+\frac{1}{2n}}}{\sqrt{1+\frac{1}{n}}+1} [/tex]

    Taking limits gives (1/2) as the answer. You can recognize the original limit as an indeterminate form and divide by the arbitrarily growing variable n to get to the same thing.

    I should mention however, that the sandwiching approach used by NateTG is far more elegant than this "trick" here (which gives you the answer but not an insight).

    Last edited: Nov 14, 2004
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