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anonymous

  • 5 years ago

lim (x^2-1)/(sqrt(x)-1) x->1

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  1. anonymous
    • 5 years ago
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    using L'Hospital rule, lim x->a ( fx/gx) = lim x-> a (f'x/g'x) so lim x->1 (x^2-1)/(sqrt(x)-1) = lim x->1 (2x)/ (1/2 x^-1/2) = 2/ 1/2 = 4

  2. anonymous
    • 5 years ago
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    but the ans in the book says 2

  3. anonymous
    • 5 years ago
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    I have not learned L'hospital's rule yet

  4. anonymous
    • 5 years ago
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    i'm pretty sure the answer is 4 if i use another way, by multiply it with sqrt(x)+1/sqrt(x)+1, i got the same anwer lim x-> 1 (x-1)(x+1) (sqrt(x)+1)/(sqrt(x)-1)(sqrt(x)+1) = lim x->1 (x-1)(x+1)(sqrt(x)+1)/(x-1) = lim x->1 (x+1)(sqrt(x) +1) = 2(1+1) = 4

  5. anonymous
    • 5 years ago
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    ok thank you very much!

  6. anonymous
    • 5 years ago
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    you're welcome

  7. amistre64
    • 5 years ago
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    f(x) = x^2 -1 --------- g(x) = sqrt(x) - 1 L'Hopital says that the limit of this function can be lim| f'/g' 2x --------- = 4x sqrt(x) = 4(1)(sqrt(1)) = 4 1/2sqrt(x) I agree ...... or messed it up :)

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