# B Continuous but Not Differentiable

1. Mar 30, 2016

### logan3

Suppose a certain function in continuous at $c$ and $(c. f(c))$ exists, then which of the two could be false: $\displaystyle \lim_{x \rightarrow c^-} {f(x)} = \lim_{x \rightarrow c^+} {f(x)}$, and $\displaystyle f'(c)$?

I feel like both could be false, because if the formal derivative at a point exists, then the left and right hand limits much be equal -- but the function could be $f(x) = |x|$, which means that $\displaystyle \lim_{x \rightarrow 0^-} {f(x)} \neq \lim_{x \rightarrow 0^+} {f(x)}$ (but $\displaystyle \lim_{x \rightarrow 0} {f(x)} = 0$) and $f'(0) = \frac {d}{dx}|0|$ does not exist. I feel like I'm missing something, cause the nuances around continuity and differentiability have always been confusing (and vague) to me.

Thank-you

2. Mar 30, 2016

### Samy_A

If f is continuous in c, then $\displaystyle \lim_{x \rightarrow c^-} {f(x)} = \lim_{x \rightarrow c^+} {f(x)} =f(c)$.

For $f(x)=|x|$, what you have is that $\displaystyle \lim_{x \rightarrow 0^-} \frac{f(x)}{x}=\lim_{x \rightarrow 0^-} \frac{-x}{x}=-1$ and $\displaystyle \lim_{x \rightarrow 0^+} \frac{f(x)}{x}=\lim_{x \rightarrow 0^+} \frac{x}{x}=1$. That's why the function has no derivative in 0.

3. Mar 30, 2016

### HallsofIvy

No, this is wrong. Both $\displaystyle \lim_{x \rightarrow 0^-} {f(x)}$ and $\lim_{x \rightarrow 0^+} {f(x)}$ exist and are equal to 0.
If $\lim_{x\to a} f(x)$ exists then it must be true that $\lim_{x\to a^-} f(x)$ and $\lim_{x\to a+} f(x)$ exist and are equal. Perhaps you are thinking of the fact that the limits of the derivatives are not equal: $\displaystyle \lim_{x \rightarrow 0^-} {f'(x)} \neq \lim_{x \rightarrow 0^+} {f'(x)}$.
(Note f'(x), not f(x).)