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Math Help - Limits question

  1. #1
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    Limits question

    Limit[x^0.5 ((x + 4)^0.5 - x^0.5), x -> ∞]

    1. does not exist
    2. exists and is equal to 0
    3. exists and is equal to 1/2
    4. exists and is equal to 2

    (The answer is given as D. But how to proove it? )
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  2. #2
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    you sure the answer isn't -2? or the limit goes to 0?
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  3. #3
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    yes. the answer is 2. (and NOT -2)
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  4. #4
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    why don't you try to graph it with a software? I am pretty sure the limit approches to -2, if there is no mistake in the paranthesis,
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  5. #5
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    yes. graphing software show that the limit appproaches 2.

    Plot[x^0.5` (-x^0.5` + (4 + x)^0.5`), {x, -10, 10}]
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  6. #6
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    oki, here is my solution.

    I multiplied and divided the term with the conjugate of the second part of the equation. then used L'hospital. Because it is inf/inf.
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  7. #7
    MHF Contributor chisigma's Avatar
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    With simple manipulations you obtain...

    \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x})= \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x}) \cdot \frac{\sqrt{x+4}+\sqrt{x}}{\sqrt{x+4}+\sqrt{x}} =

    = \frac{4\cdot \sqrt{x}}{\sqrt{x+4}+\sqrt{x}}= \frac{4}{1+\sqrt{\frac{x+4}{x}}}

    But...

    \lim_{x \rightarrow \infty} \sqrt{\frac{x+4}{x}}=1

    ... so that...

    \lim_{x \rightarrow \infty} \sqrt{x} \cdot (\sqrt{x+4}-\sqrt{x})= 2

    Kind regards

    \chi \sigma
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  8. #8
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    impressed with the use of symbols.
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  9. #9
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    Quote Originally Posted by sench View Post
    impressed with the use of symbols.
    You also can learn LaTeX
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