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Approximation of a function in limit of small and large x

  1. Aug 2, 2012 #1
    Hi all,

    I'm having a bit of trouble getting my head round approximations to a function in the limit of small and large values of the x parameter. The function is:

    [itex]y = x\left\{ {\left[ {1 + \left( {{1 \over x}} \right)^2 } \right]^{{1 \over 2}} - 1} \right\}[/itex]

    The paper I'm reading says y becomes independent of x for small values of x. For large values of x, y becomes proportional to (1/x). Has anyone got any ideas of how this can be shown algebraically? I haven't had to do this for a while!

    FD
     
  2. jcsd
  3. Aug 2, 2012 #2

    Let us look at
    [tex]h(x):=\frac{\left(1+\frac{1}{x^2}\right)^{1/2}-1}{\frac{1}{x}}[/tex]

    When [itex]\,x\to 0\,[/itex] , both numerator and denominator above go to [itex]\,\pm \infty[/itex] , so applying L'Hospital we get:

    [tex]\lim_{x\to 0}h(x)=\lim_{x\to 0}\frac{-\frac{1}{x^3}\left(1+\frac{1}{x^2}\right)^{-1/2}}{-\frac{1}{x^2}}=\lim_{x\to 0}\frac{1}{x\left(1+\frac{1}{x^2}\right)^{1/2}}=\lim_{x\to 0}\frac{1}{\left(x^2+1\right)^{1/2}}=1[/tex]

    the above being possible if [itex]\,x>0\,\,,\,i.e.\,\,\lim_{x\to 0^+}\,[/itex] . If we take the limit from the left just

    multiply by [itex]\,-1\,[/itex] after the 3rd. equality above and proceed then as before. The limit then is [itex]\,-1[/itex] .

    This shows the paper you're reading is wrong as the value approached definitely depends on
    whether x is "very small",

    positive or negative . It's easy to see that has x becomes larger and larger the expression in the square root

    becomes closer to 1.

    Doing exactly the same as above in the case [itex]\,x\to\infty\,[/itex] , we get the limit is now zero, but the

    thing about "proporcionality" perhaps means that [itex]\,|h(x)|\leq\frac{M}{x}\,\,\,,\,\,as\,\,\,x \to \infty\,\,,\,M=\,[/itex] a constant. Check this.

    DonAntonio
     
  4. Aug 2, 2012 #3

    mfb

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    You can calculate the taylor expansion for h(x) at x=0 and for h(1/x) at 1/x=0 - in both cases, use the limit value where h is not well-defined. If x cannot be negative, this limit exists.


    Hand-wavy way for small x: (1+z)^(1/2) = 1+z/2+O(z^2) for small z, therefore
    [itex]h(x) = x \left(\frac{1}{x}(1+x^2)^{1/2}-1\right) = x\left(\frac{1}{x}(1+\frac{x^2}{2}+O(x^4))-1\right)=1-x+\frac{x^2}{2}+O(x^4)[/itex]
    Well, I would not call this "independent of x", the single "-x" does not fit to this.
    Another way would be to derive your whole equation. If the limit of the derivative is 0 (it is not, unless I made a mistake), you could call this "independent of x".


    And small 1/x:
    [itex]h(x)=x \left(1+\frac{1}{x^2}+O(\frac{1}{x^4})-1\right) = \frac{1}{x} + O(\frac{1}{x^3})[/itex]
    Another way here would be to show that y*x approaches a constant for x->infinity.


    @DonAntonio: This would be an upper limit, not a proportionality.
     
  5. Aug 5, 2012 #4
    Isn't that the same as:

    [itex]f(x) = x \times \sqrt {1 + \left( {{1 \over x}} \right)^2 } - x[/itex]

    which is equal to:

    [itex]f(x) = \sqrt {x^2 + x^2 \times\left( {{1 \over x^2}} \right) } - x[/itex]

    which is equal to:

    [itex]f(x) = \sqrt {x^2 + 1 } - x[/itex]

    Which leads us to the conclusions:
    [tex]\lim_{x\to\infty} \sqrt {x^2 + 1 } - x =0[/tex]
    and
    [tex]\lim_{x\to 0} \sqrt {x^2 + 1 } - x = 1[/tex]
     
  6. Aug 5, 2012 #5

    pwsnafu

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    x can be negative so the next line is wrong. Remember [itex]\sqrt{x^2} = |x|[/itex]
     
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