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Math Help - a question on calculus -limits

  1. #1
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    Red face a question on calculus -limits

    solve

    lim ((1+2x)^0.5-(3x)^0.5)/((3+x)^0.5-2((x)^0.5))
    x→1


    Please help me!
    -Riyafa-
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  2. #2
    MHF Contributor red_dog's Avatar
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    \lim_{x\to 1}\frac{\sqrt{1+2x}-\sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}}=\lim_{x\to 1}\frac{1-x}{\sqrt{1+2x}+\sqrt{3x}}\cdot\frac{\sqrt{3+x}+2\s  qrt{x}}{3-3x}=

    =\frac{1}{3}\lim_{x\to 1}\frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+\sqrt{3x  }}=\frac{1}{3}\cdot\frac{4}{2\sqrt{3}}=\frac{2\sqr  t{3}}{9}
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  3. #3
    MHF Contributor
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    Quote Originally Posted by riyafa View Post
    solve

    lim ((1+2x)^0.5-(3x)^0.5)/((3+x)^0.5-2((x)^0.5))
    x→1


    Please help me!
    -Riyafa-
    Rationalize both top and bottom

     <br />
\lim_{x \to 1} \frac{\sqrt{1+2x}- \sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}}<br />

     <br />
=\lim_{x \to 1} \frac{\sqrt{1+2x}- \sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}} \cdot \frac{\sqrt{1+2x}+ \sqrt{3x} }{\sqrt{1+2x}+ \sqrt{3x} } \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{3+x}+2\sqrt{x} }

     <br />
=\lim_{x \to 1} \frac{\sqrt{1+2x}- \sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}} \cdot \frac{\sqrt{1+2x}+ \sqrt{3x} }{\sqrt{3+x}+2\sqrt{x} } \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+ \sqrt{3x} }

     <br />
=\lim_{x \to 1} \frac{1-x}{3 - 3x} \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+ \sqrt{3x} }

     <br />
=\lim_{x \to 1} \frac{1}{3} \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+ \sqrt{3x} } = \frac{2}{3\sqrt{3}}
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