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anonymous

  • 4 years ago

Compute A= x^2+x/x+1 as x approaches +infinity and B= x^2+x/x^2+1 as x approaches -infinity

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  1. anonymous
    • 4 years ago
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    By using l'hopital's rule, take the derivative of \[x ^{2}+x = 2x+1\] and the derivative of \[x+1 = 1\] \[A = +\infty\] Similary for B = 1

  2. anonymous
    • 4 years ago
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    im not familiar with the l'hopital rule

  3. anonymous
    • 4 years ago
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    is there a way i can do it geometrically

  4. anonymous
    • 4 years ago
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    I don't think so. L'hopital's rule states that if you plug in the infinity into that equation and when the equations gives \[\infty/\infty\] or 0/0 you take the derivative of the equations separetly.

  5. anonymous
    • 4 years ago
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    since im not familiar with this rule, can you walk me through it

  6. anonymous
    • 4 years ago
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    Did you read my post above? I gave you a little info about the rule. If you need detailed help just tell me.

  7. anonymous
    • 4 years ago
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    yes please, i need detailed help

  8. anonymous
    • 4 years ago
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    For example take the equation above \[A = x ^{2}+x/x+1\] If you replace +infinity in the all of the x-s you get infinity/infinity which is an indeterminate form i.e. you cannot compute the answer. Because it is an indeterminate form you have to take the derivative of those equation seperatly, like i did in my first post.

  9. anonymous
    • 4 years ago
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    ok so what happens to the 2x+1/1

  10. anonymous
    • 4 years ago
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    replace infinity in that x which gives\[2(\infty)+1/1 = \infty\]

  11. anonymous
    • 4 years ago
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    oh ok i see

  12. anonymous
    • 4 years ago
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    so B= 2x+1/2x+1

  13. anonymous
    • 4 years ago
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    2(-inf)+1/2(-inf)+1 = -inf?

  14. anonymous
    • 4 years ago
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    No. The derivative of 1 = 0. The derivative of any constant is equal to 0

  15. anonymous
    • 4 years ago
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    i understand now thanks

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