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November 28, 2014

November 28, 2014

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

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 11:49pmdivision by zero is NOT allowed

(x+5)/(x-3) is valid for all x except x=3

- READ -
**READ**, Sunday, November 1, 2009 at 12:15ambut I can evalute using limits...

- CAN SOMEONE READ THIS -
**CAN SOMEONE READ THIS**, Sunday, November 1, 2009 at 7:15amplease

- Calculus -
**drwls**, Sunday, November 1, 2009 at 9:36amWhatever you did using limits is wrong.

- Calculus -
**MathMate**, Sunday, November 1, 2009 at 10:07amThe 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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