Expert: Alon Mandes - 10/15/2008

Question
I'm struggling with grasping on how to get the answer to this question. The book shows no examples and we didn't even go over it in class, but he assigns it in our work. the back of the book shows answers but not how to get them. that is my main problem.
an example is:

Find all values x = a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit does not exist.

f(x) = 5+x
     x(x-2)
thanks for your help.

Answer
I Didn't understand if the function that you wrote is 5+x or
(5+x)/x(x-2), but I will solve for both case.

Case 1
f(x)=5-x or f(x)=x(x-2) is a continues Polynom with no singular
(discontinues) points, & the limit when x->a is clearly substituting f(a).

Case 2
f(x)=(x+5)/x(x-2) has two singular (discontinues point): 0 & -2.
In these points the function simply jumps rapidly to Infinity.
To prove so, we must go back to the Mathematical definition of limits.
Let's now prove that when x approaches 2 the function f goes to ∞ :
Lim {x->2} (x+5)/x(x-2) = (7/2) * Lim {x->2} 1/(x-2)
Lim {x->2} 1/(x-2) = ∞ means :
For each real M > 0 (very big) there exists a real ε > 0 such that for all x with 0 < |x − 2| < ε , we have |f(x)| > M.
|1/(x-2)|>M that means |x-2|<1/M . So ε=1/M . Q.E.D
If 1/(x-2) -> ∞ then (7/2)/(x-2) -> ∞ as well !!

I will leave x->0 for you as an exercise..

Alon.