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Math Help - [SOLVED] Find a limit

  1. #1
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    [SOLVED] Find a limit

    I don't understand how to simplify after multiplying the numerator and denominator by the conjugate.

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  2. #2
    Super Member Gamma's Avatar
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    That is the beauty of multiplying by the conjugate, the cross terms always cancel out.

    (a+b)(a-b)=a^2-ab+ab-b^2=a^2-b^2.

    In this case you have (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=\sqrt{a}^2-\sqrt{ab}+\sqrt{ab}-\sqrt{b}^2=\sqrt{a}^2-\sqrt{b}^2=a-b

    Where a=\frac{x+\Delta x}{5} and b=\frac{x}{5}. So a-b=\frac{x+\Delta x}{5}-\frac{x}{5}=\frac{x+\Delta x-x}{5}=\frac{\Delta x}{5}
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  3. #3
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    ok, so now how do I go from

    \frac{\frac{\triangle x}{5}}{\triangle x (\sqrt{\frac{x+\triangle x}{5}}+\sqrt{\frac{x}{5}})}

    to

    \frac{1}{\sqrt{5}(\sqrt{x+\triangle x}+\sqrt{x})}
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  4. #4
    Super Member Gamma's Avatar
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    Quote Originally Posted by Mike9182 View Post
    ok, so now how do I go from

    \frac{\frac{\triangle x}{5}}{\triangle x (\sqrt{\frac{x+\triangle x}{5}}+\sqrt{\frac{x}{5}})}

    to

    \frac{1}{\sqrt{5}(\sqrt{x+\triangle x}+\sqrt{x})}


    Cancel the \Delta x and move the 5 to the denominator, also go ahead and factor out a \frac{1}{\sqrt5} from that fraction
    \frac{\frac{\triangle x}{5}}{\triangle x (\sqrt{\frac{x+\triangle x}{5}}+\sqrt{\frac{x}{5}})}=\frac{1}{5\frac{1}{\sq  rt{5}}(\sqrt{x+\Delta x} +\sqrt{x})}= \frac{1}{\sqrt{5}^2\frac{1}{\sqrt{5}}(\sqrt{x+\Del  ta x} +\sqrt{x})}=\frac{1}{\sqrt{5}(\sqrt{x+\Delta x} +\sqrt{x})}

    Now take the limit as \Delta x \rightarrow 0 and that term goes away and you get
    \frac{1}{\sqrt{5}(\sqrt{x+\Delta x} +\sqrt{x})}=\frac{1}{\sqrt{5}(2\sqrt{x})}=\frac{1}  {2\sqrt{5x}}=\frac{1}{\sqrt{2^2}\sqrt{5x}}=\frac{1  }{\sqrt{20x}}
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