Homework Help Forum: Calculus

Posted by Calculus on Saturday, October 31, 2009 at 11:45pm.

can someone please explain h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 5 7 0 4 1 1 4 5

• Calculus - bobpursley, Saturday, October 31, 2009 at 11:49pm

  division by zero is NOT allowed (x+5)/(x-3) is valid for all x except x=3

  
  

• READ - READ, Sunday, November 1, 2009 at 12:15am

  but I can evalute using limits...

  
  

• CAN SOMEONE READ THIS - CAN SOMEONE READ THIS, Sunday, November 1, 2009 at 7:15am

  please

  
  

• Calculus - drwls, Sunday, November 1, 2009 at 9:36am

  Whatever you did using limits is wrong.

  
  

• Calculus - MathMate, Sunday, November 1, 2009 at 10:07am

  The limit at x=3 exists if lim x→3- equals lim x→3+. In this case, the two are not equal, because they tend toward inifinity in opposite directions, so the limit does NOT exist. Even if they tend toward infinity in the same direction, the limit does not exist. Even in the case where the limit exists, as in the case of f(x)=(x-3)*(x+5)/(x-3), the point x=3 remains a hole in the domain of f(x). In the case of f(x)=(x+5)/(x-3), there is no doubt that the point x=3 is excluded from the domain, which is ℝ\3.

  
  

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