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September 30, 2014

September 30, 2014

Posted by **Terry** on Friday, September 3, 2010 at 1:53pm.

1. f(x)= (x^4-1)/(x-1), a=1

?The discontinuity is removable.

?The discontinuity is not removable.

g(x) = ?

2. f(x)= (x^3-x^2-12x)/((x-4), a=4

?The discontinuity is removable.

?The discontinuity is not removable.

g(x) = ?

3) f(x) = [[ sin(x) ]], a = ð

(Recall that [[h(x)]] means the largest integer that is less than or equal to h(x).)

?The discontinuity is removable.

?The discontinuity is not removable.

g(x) =?

- calculus please help! -
**Terry**, Friday, September 3, 2010 at 2:08pmI dont understand what they are asking! If anyone could explain that, that would be a great help!

- calculus please help! -
**MathMate**, Friday, September 3, 2010 at 3:46pmA removable discontinuity is a point where the function does not exist, but it's limit exists on both sides for an interior point, and on the interior side if it is an end point.

The discontinuity can be removed by defining the function at the point of discontinuity as the limit.

Take for an example the first question:

f(x)=(x^4-1)/(x-1)

Since

x^4-1 = (x²+1)(x+1)(x-1)

it is evident that the limit at x=1 exists on both sides, but f(x) is undefined at x=1.

By redefining f(x) as

g(x)=(x^4-1)/(x-1) for x≠1, and

g(x)=4 for x=1

g(x) is now continuous on ℝ, and the discontinuity has been removed in the new function.

You can work on the other problems on this basis.

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