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Thread: Limit Calculation

  1. #1
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    Limit Calculation

    Hi guys, I need a little help with this limit here:

    lim x->infinity of: sqrt(x^2 + x) - sqrt(x^2 - 2x)

    cheers !
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  2. #2
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    Quote Originally Posted by WeeG View Post
    Hi guys, I need a little help with this limit here:

    lim x->infinity of: sqrt(x^2 + x) - sqrt(x^2 - 2x)

    cheers !
    "Rationalize the numerator"- that is, multiply both numerator and denominator by $\displaystyle \sqrt{x^2+ x}+ \sqrt{x^2- 2x}$ to get $\displaystyle \frac{(\sqrt{x^2+ x}- \sqrt{x^2- 2x})(\sqrt{x^2+ x}+ \sqrt{x^2- 2x})}{\sqrt{x^2+ x}+ \sqrt{x^2- 2x}}$$\displaystyle = \frac{x^2+ x- x^2+ 2x}{\sqrt{x^2+x}+ \sqrt{x^2- 2x}}= \frac{3x}{\sqrt{x^2+ x}+ \sqrt{x^2- 2x}}$
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  3. #3
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    done it, cheers !
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  4. #4
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    Or we can solve it by mean value theorem ,

    Let $\displaystyle f(x) = \sqrt{t} $
    we have

    $\displaystyle \frac{ \sqrt{x^2 + x } - \sqrt{x^2 - 2x}}{x^2 + x - x^2 + 2x} = \frac{1}{2\sqrt{c}} $ where $\displaystyle x^2 - 2x < c < x^2 + x$

    Take reciporcal and square root ,

    $\displaystyle \frac{1}{\sqrt{x^2 - 2x}} > \frac{1}{\sqrt{c}} > \frac{1}{\sqrt{x^2 +x }}$

    $\displaystyle \frac{1}{\sqrt{x^2 - 2x}} > 2 \frac{ \sqrt{x^2 + x } - \sqrt{x^2 - 2x}}{3x}> \frac{1}{\sqrt{x^2 +x }} $


    $\displaystyle \frac{3}{2} \frac{x}{\sqrt{x^2 - 2x}} > \sqrt{x^2 + x } - \sqrt{x^2 - 2x} > \frac{3}{2} \frac{x}{\sqrt{x^2 +x }} $

    $\displaystyle \lim_{x\to\infty} \frac{3}{2} \frac{x}{\sqrt{x^2 - 2x}}> \lim_{x\to\infty} \sqrt{x^2 + x } - \sqrt{x^2 - 2x} > \lim_{x\to\infty} \frac{3}{2} \frac{x}{\sqrt{x^2 +x }} $

    $\displaystyle \frac{3}{2} > \lim_{x\to\infty} \sqrt{x^2 + x } - \sqrt{x^2 - 2x} >\frac{3}{2} $


    $\displaystyle L = \frac{3}{2} $
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