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anonymous

  • 5 years ago

use the properties of limits to fin the indicated limit: lim x->5- x^2+25/x+5

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  1. anonymous
    • 5 years ago
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    as x goes to 5 from the left, there is no division by zero, so why not substitute?

  2. anonymous
    • 5 years ago
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    Is it really x^2+25/x^2+5?

  3. anonymous
    • 5 years ago
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    yes

  4. anonymous
    • 5 years ago
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    with lim of x->5-

  5. anonymous
    • 5 years ago
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    I still don't see the issue, unless it is really the limit as x goes to negative 5 of (x^2 + 25)/(x + 5) otherwise, you can substitute directly. If it is this, though...

  6. anonymous
    • 5 years ago
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    ooh ok thank you

  7. anonymous
    • 5 years ago
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    The only problem with simply plugging the value of x in would be is if that would cause the denominator to go to zero.

  8. anonymous
    • 5 years ago
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    Oh ok :)

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spraguer (Moderator)
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