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Confused about finding limit

  1. Nov 10, 2004 #1

    quasar987

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    I am asked to evaluate this limit

    [tex]\lim_{x\rightarrow \infty}\frac{\sqrt{3x^2+11x+7}+\sqrt{3x^2-11x+7}}{x+\sqrt{x^2+7}} [/tex]

    but without use of the basic operations on limits (ex: the limit of a sum is the sum of the limits if they exist, etc.) (because they are to be introduced only in the next chapter). I know the limit is very easily computed: just divide the numerator and the denominator by x and the answer is [itex]\sqrt{3}[/itex]:

    [tex]= \lim_{x\rightarrow \infty}\frac{\sqrt{3+\frac{11}{x}+\frac{7}{x^2}}+\sqrt{3-\frac{11}{x}+\frac{7}{x^2}}}{1+\sqrt{1+\frac{7}{x^2}}} = \sqrt{3}[/tex]

    But what is the "method" to arrive to this conclusion? The facts that

    [tex]\lim_{x\rightarrow \infty}\frac{11}{x} = 0[/tex]

    and that

    [tex]\lim_{x\rightarrow \infty}\frac{7}{x^2} = 0[/tex]

    are useless if we cannot use the operations on limits. I hope I made clear where my confusion comes from. Thanks.
     
  2. jcsd
  3. Nov 10, 2004 #2

    kreil

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    There are 3 rules of infinite limits:

    1. If the power of the top is the same as the power of the bottom, take the coefficients:

    [tex]
    \lim_{x\rightarrow \infty} \frac {4x^2 + 3x + 7}{2x^2 + 2x + 1} = \frac {4}{2}=2
    [/tex]

    2. If the power of the bottom is larger than the power of the top, the limit is zero:

    [tex]
    \lim_{x\rightarrow \infty} \frac {x^2 + 4x}{x^3 + 7x} = 0
    [/tex]


    3. If the power of the top is larger than the power of the bottom, the limit does not exist:

    [tex]
    \lim_{x\rightarrow \infty} \frac {x^3 + 3x - 7}{x^2 + 4x} = dne
    [/tex]

    I don't think these are operations; they are just facts. In your case:

    [tex]
    \lim_{x\rightarrow \infty}\frac{\sqrt{3x^2+11x+7}+\sqrt{3x^2-11x+7}}{x+\sqrt{x^2+7}}
    [/tex]

    the top and bottom are only to the power [tex] x^1 [/tex] and so you take the coefficients, [tex] \frac {\sqrt{3}}{1}=\sqrt{3} [/tex] Hopefully this helps you.
     
  4. Nov 11, 2004 #3

    quasar987

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    I'm afraid I'm not allowed to use your rules. Only the definition of limit, the caracterisation with sequences and a few theorems about the unicity and the "limit to the right, limit to the left" caracterisation.

    Let me ask the question another way: What are the steps between

    [tex]\lim_{x\rightarrow \infty}\frac{\sqrt{3+\frac{11}{x}+\frac{7}{x^2}}+\sqrt{3-\frac{11}{x}+\frac{7}{x^2}}}{1+\sqrt{1+\frac{7}{x^ 2}}}[/tex]

    and

    [tex] = \sqrt{3}[/tex]

    I tought of something this morning, could it be it? Now that we suspect using our "sense" that the limit is [itex]\sqrt{3}[/itex], we may use the [itex]\epsilon-\delta[/itex] definition to prove that it is indeed [itex]\sqrt{3}[/itex]. But I don't really see how. :shy:
     
  5. Nov 11, 2004 #4

    sal

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    Right, but because you're taking the limit as x goes to infinity, not zero, I would tend to use "N" rather than epsilon.

    To save space, let

    [tex]f(x) = \frac{\sqrt{3+\frac{11}{x}+\frac{7}{x^2}}+
    \sqrt{3-\frac{11}{x}+\frac{7}{x^2}}}
    {1+\sqrt{1+\frac{7}{x^ 2}}}
    [/tex]

    To apply the definition directly, what you need to do is assume you have a value for delta, and then solve for N in the equation:

    [tex]f(N) - \sqrt{3} = \delta[/tex]

    And once you've done that, you need to show that

    [tex]x > N \Rightarrow |f(x) - \sqrt{3}| < |f(N) - \sqrt{3}|[/tex]

    It looks likely to get messy, which, if nothing else, illustrates why we don't generally work directly from the definition of the limit.
     
  6. Nov 11, 2004 #5

    quasar987

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    Great, thanks.
     
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