# Limits and functions

1. Nov 4, 2013

### houssamxd

removable discontinuity

1. The problem statement, all variables and given/known data

the following function
f(x)=(4-x)/(16-x^2)

is discontinuous at???

i got at -4 but some of my friends say its 4, -4

how is that possible

Last edited: Nov 4, 2013
2. Nov 4, 2013

### jbunniii

Is the function even defined at $x = \pm 4$?

3. Nov 4, 2013

### houssamxd

well thats what u have to find

where is it discontinuous
i have an exam on this but its not clear yet to me

4. Nov 4, 2013

### jbunniii

Discontinuous doesn't mean the same thing as undefined. Go back to the definition of continuity: can a function be continuous at a point if it is not defined at that point?

5. Nov 4, 2013

### houssamxd

well basically we just have to use the function and find the points where its disconitinuous
i.e in the graph there is a hole or a jump

check here
http://www.dummies.com/how-to/content/how-to-determine-whether-a-function-is-discontinuo.html

6. Nov 4, 2013

7. Nov 4, 2013

### houssamxd

0

but if u factorize the denominator and cancel you will only get -4 as a point of discontinuity

try it yourself

Last edited: Nov 4, 2013
8. Nov 4, 2013

### HallsofIvy

Staff Emeritus
Are you saying that the difficulty is that you do not know what "continuous" means? Didn't It occur to you to look up the definition?

A function, f, is continuous at x= a if and only if:
(1) f(a) exists
(2) $\lim_{x\to a} f(x)$ exists
(3) [itex]\lim_{x\to a} f(x)= f(a).

For what values of x is at least one of those NOT true? (Look specifically at (1)!)

9. Nov 4, 2013

### houssamxd

but if u factorise it and cancel the common
you only get -4 as the point of discontinuity

10. Nov 4, 2013

### jbunniii

Note that
$$\frac{4-x}{16-x^2}$$
and
$$\frac{1}{4+x}$$
are not the same function. They are equal for $x \neq 4$, but the first function is undefined (hence discontinuous) at $x=4$ whereas the second is defined and continuous at $x=4$.

Therefore the first function has what kind of discontinuity at $x=4$?

11. Nov 4, 2013

### jbunniii

12. Nov 4, 2013

### houssamxd

i guess it has a removable one right??

13. Nov 4, 2013

### jbunniii

That's right. After you remove it by canceling the common factor in the numerator and denominator, the result (second function I listed above) is continuous at that point.

Now what about $x = -4$? You correctly determined that it is discontinuous there. What kind of discontinuity is it?

14. Nov 4, 2013

### houssamxd

non removable
but do they both count as discontinuities

15. Nov 4, 2013

### jbunniii

Sure. A removable discontinuity is still a discontinuity. As your link describes it, you can think of it as a "gap" in the graph of the function at $x = 4$. You can fill the gap by removing the common factor so the denominator will be defined at $x=4$.

16. Nov 4, 2013

### houssamxd

thanx alot man
i owe you one
wish me luck for tomorrows exam

17. Nov 4, 2013

### jbunniii

Absolutely, good luck!