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Math Help - need help on calculating a limit.

  1. #1
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    need help on calculating a limit.

    lim [sqrt(1+x)-sqrt(x-1)]
    x->infinity

    i tried to move it to the denominator like this:
    lim 1/[sqrt(x+1)+sqrt(x-1)]

    but here im stuck, i think it should be zero, but hoe do i prove it?

    thanks in advance.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DangerMan
    lim [sqrt(1+x)-sqrt(x-1)]
    x->infinity

    i tried to move it to the denominator like this:
    lim 1/[sqrt(x+1)+sqrt(x-1)]

    but here im stuck, i think it should be zero, but hoe do i prove it?

    thanks in advance.
    A simple (nonrigorous) way to do it is to look at \sqrt{x \pm 1}. When x is large, the 1 is so small as to be ignorable. Thus
    \lim_{x \to \infty} \left ( \sqrt{1+x} - \sqrt{x-1} \right ) \approx \lim_{x \to \infty} \left ( \sqrt{x} - \sqrt{x} \right ) \to 0

    -Dan
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  3. #3
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    but i asked for a proof, do you know of a rigorous way to prove it?

    thanks.
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  4. #4
    TD!
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    Depends on "how much rigour" you want...
    You could always go back to the definition, but I wouldn't do that unless necessary.

    Multiply numerator and denominator with the complement of the numerator to get:

    <br />
\frac{{\left( {\sqrt {x + 1}  - \sqrt {x - 1} } \right)\left( {\sqrt {x + 1}  + \sqrt {x - 1} } \right)}}{{\left( {\sqrt {x + 1}  + \sqrt {x - 1} } \right)}} = \frac{2}{{\left( {\sqrt {x + 1}  + \sqrt {x - 1} } \right)}}<br />

    Now the numerator is constant and positive and the denominator clearly goes to +infinity, yielding 0 as limit.
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  5. #5
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    You can do even this,
    note that for x\geq 1 we have,
    0\leq \sqrt{x+1}-\sqrt{x-1} \leq \frac{1}{\sqrt{.5x}}
    But,
    \lim_{x\to\infty} 0=0 and,
    \lim_{x\to\infty} \frac{1}{\sqrt{.5 x}}=0

    Thus, you have a function,
    f(x)=\sqrt{x+1}-\sqrt{x-1}
    squeezed between two functions which have the same limit. Conclude from the squeeze theorem that,
    \lim_{x\to\infty}\sqrt{x+1}-\sqrt{x-1}=0
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