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anonymous

  • 4 years ago

[(e^x)-(1)-(x)]/ (x^2) as lim x --> 0

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  1. anonymous
    • 4 years ago
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    Again apply L' Hospital's Rule to get 1/2 as the limit.

  2. anonymous
    • 4 years ago
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    how? =(

  3. anonymous
    • 4 years ago
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    expand e^x and see it fall all in place ans is 1/2

  4. anonymous
    • 4 years ago
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    cant get the answer but i believe you

  5. anonymous
    • 4 years ago
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    lol

  6. anonymous
    • 4 years ago
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    how exactly did u expand e to x

  7. anonymous
    • 4 years ago
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    The given limit is in its indeterminate form thus we differentiate the numerator and denominator to get 1/2[(e^x - 1)/x] now the limit of [e^x - 1/]x is nothing but 1.

  8. anonymous
    • 4 years ago
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    e^x is expanded by using the Taylor Series.

  9. anonymous
    • 4 years ago
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    That is euler series not Taylor

  10. anonymous
    • 4 years ago
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    i didnt know you can just differentiate the numerator and the denominator while the numerator was being subtracted from e^x - 1

  11. anonymous
    • 4 years ago
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    It is actually Maclaurin series which is a special case of Taylor Series.

  12. anonymous
    • 4 years ago
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    If you want you can actually prove it using cauchy mean valuue theorem

  13. anonymous
    • 4 years ago
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    can someone quickly show me how you get the answer to my other problem under this? thank you

  14. anonymous
    • 4 years ago
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    WHere is the other problem

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