Let f be a continuous real function on R, of which it is known that f'(x) exists for all nonzero x, and that f'(x) -> 3 as x -> 0. Does it follow that f'(0) exists?
Solution:
\( \displaylines{f\left(x+h\right)-f\left(x\right)=hf^{\prime}\left(c\right)=h\cdot3\text{ for some }c\in\left\lbrack x,h\right\rbrack.\\ \text{As f is continuous, the difference }f\left(h\right)-f\left(0\right)\text{ is the limit of LHS as }x\to0\\ \text{Thus}f\left(h\right)-f\left(0\right)=hf^{\prime}\left(c\right)\text{ for some }c\in\left\lbrack0,h\right\rbrack\\ \text{Letting }h\to0\text{ indicates }f^{\prime}\left(0\right)=3} \)