anonymous
  • anonymous
use the properties of limits to fin the indicated limit: lim x->5- x^2+25/x+5
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
as x goes to 5 from the left, there is no division by zero, so why not substitute?
anonymous
  • anonymous
Is it really x^2+25/x^2+5?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
with lim of x->5-
anonymous
  • anonymous
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...
anonymous
  • anonymous
ooh ok thank you
anonymous
  • anonymous
The only problem with simply plugging the value of x in would be is if that would cause the denominator to go to zero.
anonymous
  • anonymous
Oh ok :)

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