Homework Help: 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

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!