(Non)removable discontinuity?

Question

I'm a little confused on the subject of nonremovable vs. removable discontinuity. I know removable means you could make a graph continuous by substituting in a different value, but does that apply only when a y value exists but just isn't continuous?

For example...if you had function that simplified to:

(x-2) / (x-2)(x+2)

You would factor out (x-2) and have discontinuity at 2. But since no y value for x=2 exists at all, is it removable or nonremovable?

Answer 1

Check out this question. It's very similar to yours.
http://answers.yahoo.com/question/index;_ylt=ArK7ouYwyp9DyLJRzTX.bw_ty6IX;_ylv=3?qid=20070920103416AAv5eWm

There is a removable discontinuity at x=2. Your function is identical to y = 1/(x+2) except that there is a hole at x = 2. Removable discontinuity means that if you specify y(2) = 1/4, you can make the function continuous.

Answer 2

If you divide out the (x-2) factor you have 1/(x+2), and when x = 2, its value is 1/4. The limit of (x-2)/(x²-4) as x→2 is 0/0, so using L'Hopital, take derivatives of top and bottom, getting 1/2x, and as x→2, that → 1/4. This is a removable discontinuity, a simple hole in the graph that can be handled by defining f(2) = 1/4. The vertical asymptote at x = -2 is a discontinuity that cannot be removed.

Answer 3

for ( x - 2) / ( x - 2)(x + 2), you would have a removable discontinuity at x = 2.

Why ?
by taking limits, you get a TI [ temp. indeterminate ] ratio of 0 / 0, and the ( x - 2) terms cancel out, leaving 1 / ( x + 2) ... even though the function is undefined at x = 2, you can re-define the function so that at x = 2, the y value is = 1 ( 2 + 2 ) = 1 / 4 , thus 'repairing' the hole in the graph ...

[[ the limit of the function as x -->2 is 1 / 4 ]]

Now x = - 2 is non removable, since it does not "cancel out" like the other term did... e.g, you end up with 1 / ( x + 2), and you can not repair this function at this value...

The graph of this function is a hyperbola... with a vertical asymptote at x = -2, and a missing 'point' at x = 2, [ hard to find, but graphing on a Graphic calculator and zooming in a few times usually will show the hole ], but the missing point can be repaired... think of the single missing point like a pinhole in a bicycle tire... it can be repaired, while the point at x = -2 is like a huge tear in the tire... it can't be repaired...

Answer 4

Yup, its D. (x^2 - 9) components to (x + 3)(x - 3). Then the (x + 3) on the best cancels out the (x + 3) on the backside, making x = -3 the detachable discontinuity. so which you're left with a million/(x - 3), which skill x = 3 is a vertical asymptote, and hence a nonremovable discontinuity.