Testing Limits: Can False Statements Exist?

[Anonymous217]

Homework Statement

If $$\lim_{x\rightarrow c} f(x) = f(c)$$ for all values of c, 0<or= c <or=5, and f(0)$$\neq$$f(5), which of the following could be false?
A. f(4) exists

B. f'(1) exists

C. $$\lim_{x\rightarrow2^+} f(x)$$ exists

D. $$\lim_{x\rightarrow3} f(x) = \lim_{x\rightarrow3^+} f(x)$$

E. $$\lim_{x\rightarrow0^+} f(x) \neq \lim_{x\rightarrow5} f(x)$$

The Attempt at a Solution

I tried process of elimination, but they all seemed true.
A. f(4) exists simply by the first condition.
B. f'(1) should exist because the limit for f(1) exists.
C. this should exist if limit of x to 2 f(x) exists.
D. If the limit on both sides exist, it should be equal to the right side or the left side as well.
E. limit of 0 from the right side = limit of 0 from both sides, which is = f(0); the other part is = f(5), so this statement should be true.

Clearly I've made a mistake in my logic somewhere.. :(EDIT: Oh, wait, could it be B? If you do a cusp at x=1, then the f(1) exists, but f'(1) doesn't.

 

[sutupidmath]

It looks to me B. Because, eventhough f, with the given properties/conditions, is continuous at 1, remember that continuity does not imply differentiability(while the converse is true).

 

[Altabeh]

EDIT: Oh, wait, could it be B? If you do a cusp at x=1, then the f(1) exists, but f'(1) doesn't.

You got it right! For instance, $$f(x)={\frac {\sqrt { \left| x \right| }}{{x}^{2}+1}$$ vanishes at x=0; while its derivative $$1/2\,{\frac { \left| 1 \right| }{\sqrt { \left| x \right| } \left( {x}^{2}+1 \right) }}-2\,{\frac {\sqrt { \left| x \right| }x}{ \left( {x}^{2}+1 \right) ^{2}}}$$ is not defined at the same point.

AB

 

[Anonymous217]

Thanks for the confirmation guys.