find the limit of (sqrt(2-x)-sqrt(2))/x as x->0?

Without using L'Hostpital's rule...

This is asked as a simple question in introductory calculus. I have done loads of calculus and am starting to feel stupid that I cannot get this. I have tried (to the extent of wasting trees) and googled/wolfram'd the answer. I know the answer is -1/2(sqrt2) but its the why that is annoying me! Wolfram uses L'Hostpital's rule... I'm sure it's reeeeally simple?

Help?

Update:

ah! Conjugate, then apply limit laws as per usual... right... got it... obviously I'm going mad =D

2 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    (lim x → 0) (√(2 - x) - √2)/x

    = (√(2 - x) - √2)(√(2 - x) + √2) / (x (√(2 - x) + √2))

    = ((2 - x) - 2) / (x (√(2 - x) + √2))

    = -x / (x (√(2 - x) + √2))

    = -1 / (√(2 - x) + √2)) Applying limit

    = -1 / (√(2 - 0) + √2)

    = -1 / (2√2)

    = (-1/2)√2.

    No, my answer is right: http://www.wolframalpha.com/input/?i=limit+%28%28s...

  • nle
    Lv 7
    1 decade ago

    the conjugate quantity is sqrt(2-x) + sqrt(2)

    then multiply and divide by the conjugate quantity.

    By the way the answer is - (1/4) * sqrt(2)

    You should put the parentheses in your answer

    -1/2(sqrt2) should be -1/ [ 2 sqrt(2) ]

    which can be simplified to my answer - (1/4) * sqrt(2)

    or - sqrt(2) /4 to avoid confusion.

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