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Sequence Limits: Difference of radicals

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the limit of the following sequence:

    2. Relevant equations
    [itex] \lim_{n \rightarrow + \infty} \sqrt[4] {2n + 1} - \sqrt[4] {n + 1} [/itex]

    3. The attempt at a solution
    I've tried multiplying the first radical by ## \frac{ \sqrt[4] {2n - 1} } { \sqrt[4] {2n - 1} } ## to make the radical into a square root (and do the analogous thing for the second radical), but that seems to lead nowhere as well as give me an extra denominator to work with.

    I've tried multiplying the second radical by 16/16 to get a 2 inside the radical, but that leaves me with ## \sqrt[4] {2n - 2} ##, which isn't much better.

    I've tried eyeballing the solution as ##n## approaches infinity; both radicals approach infinity, but ## \infty - \infty ## is indeterminate, and I think I'm supposed to solve it without L'Hospital's rule (it was in my precalculus exercise book before derivatives and L'Hospital's).

    I've got about 6 more exercises like this one, with increasingly complex polynomials under radicals.

    What am I supposed to do to simplify the exercise? Get both terms under the same radical?

    P.S. Sorry if I messed up the limit syntax, this is my first time with LaTeX...
     
  2. jcsd
  3. Oct 18, 2014 #2

    PeroK

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    Can you see what the answer is first? Before you try to prove it.
     
  4. Oct 18, 2014 #3
    Nope, I'm afraid not. My best guess is ## \sqrt[4] {n} ## as ##n## approaches infinity, but that's not a valid answer, is it?
     
  5. Oct 18, 2014 #4

    HallsofIvy

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    Multiplying by [itex]\frac{\sqrt[4]{2n- 1}}{\sqrt[4]{2n- 1}}[/itex] will give a square root in the denominator but leave a fourth root in the denominator so I don't see that as helping.

    Use, rather, the fact that [itex]x^4- y^4= (x- y)(x^3+ x^2y+ xy^2+ y^3[/itex] with [itex]x= \sqrt[4]{2n+1}[/itex] and [itex]y= \sqrt[4]{n+ 1}[/itex].
     
  6. Oct 18, 2014 #5

    PeroK

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    You should be able to see the answer by using some estimation. Think about the two terms and try to estimate them in ways that makes things simpler.

    (And, why not plug the largest value of n you can into a calculator or spreadsheet? Just to see!)

    PS Your best guess is not quite right, but it may still give you an idea of how to manipulate the expression.
     
    Last edited: Oct 18, 2014
  7. Oct 18, 2014 #6
    I've used that formula, and another one; the result is a rather pretty numerator of n... over a rather ugly denominator of sums of products of roots of n's and 2n's. I don't know what to do with it.

    I've checked with very large numbers and a graphing calculator and saw that the function is technically increasing... veeery slowly, but it is increasing apparently to infinity. If it were plotted in x-y coordinates, you'd get points with an x of 1 billion and a y of... 33, if I remember well. I can't figure out the limit intuitively.
     
  8. Oct 18, 2014 #7

    PeroK

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    I reckon you need a bit of help. Here's how I looked at it:

    ##\sqrt[4] {2n + 1}## tends to ##\sqrt[4] {2n}## and ##\sqrt[4] {n + 1}## tends to ##\sqrt[4] {n}##

    So, the difference tends to ##(\sqrt[4] {2}-1)\sqrt[4] {n}## and hence the sequence diverges.

    This informal analysis should also give you the clue to take out a factor of ##\sqrt[4] {n}## in order to show this formally.
     
  9. Oct 18, 2014 #8
    The first line of this argument has been quite obvious to me from the beginning... yet somehow I never made the jump to the eventual conclusion. Now it all makes sense. Thanks a lot!
     
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