Confusion about continuity question.

mathstudent79
Messages
9
Reaction score
0
1.The Question

The function f(x)=


x2/x if (x≠0)



0 if(x=0)

The Attempt at a Solution

I thought this had a removable discontinuity at x=0, because the function s2/x is not defined at x=0, and the expression can be reduced to x.

The book (one of those AP prep guides with solutions) says that it is continuous everywhere, to note that x2/x if x≠0 and lim(x->0) f = 0.

Why was I wrong?

Thanks in advance.
 
Physics news on Phys.org
mathstudent79 said:
1.The Question

The function f(x)=

x2/x if (x≠0)

0 if(x=0)

The Attempt at a Solution



I thought this had a removable discontinuity at x=0, because the function s2/x is not defined at x=0, and the expression can be reduced to x.

The book (one of those AP prep guides with solutions) says that it is continuous everywhere, to note that x2/x if x≠0 and lim(x->0) f = 0.

Why was I wrong?

Thanks in advance.
The function g(x) = x2/x does have a removable discontinuity at x=0.

The piecewise function, f(x) that you give has that discontinuity removed. f(x) is defined for all x, and is continuous for all x, even for x = 0.
 
SammyS

Thanks so much for your quick response.


So, just to be sure, once the discontinuity is removed, the function is continuous everywhere, is that right?

Thanks again.
 
mathstudent79 said:
SammyS

Thanks so much for your quick response.


So, just to be sure, once the discontinuity is removed, the function is continuous everywhere, is that right?

Thanks again.

Yes, that's right.
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top