0
$\begingroup$

Assume $f$ is differentiable at $x$,

$$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} .$$

which can be proven by,

$$ \begin{eqnarray*} \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} &=& \frac12 \lim_{h\to 0}\left(\frac{f(x+h)-f(x)}h+\frac{f(x)-f(x-h)}h\right) \\ &=& \frac12 (f'(x)+f'(x)) = f'(x) \end{eqnarray*} $$

but here's the real question,

Give an example of a function $f(x)$ for which the limit above exists at some $x$, but $f$ is not differentiable at $x$.

I am confused by this question because I originally, thought, by definition of differentiation, existence of limit at some $x$ guarantees differentiability at that 'some' $x$

$\endgroup$

4 Answers 4

Reset to default
4
$\begingroup$

Take $x=0$ and make $f$ an even function. Then we have $$\frac{f(x+h)-f(x-h)}{2h}=\frac{f(h)-f(-h)}{2h}=0$$ for all $h$ and so $$\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}=0\ .$$ It remains to find an example where $f$ is not differentiable at $0$. A somewhat trivial example is $$f(x)=\cos\frac{1}{x}$$ - this is not differentiable at $0$ because it is not defined at $0$. (But the limit above does exist - think carefully about the definition of limit.) A more interesting example: $$f(x)=|x|$$ which is continuous but not differentiable at $0$. I leave you to check the details.

$\endgroup$
1
  • $\begingroup$ couldnt ask for more, thank you $\endgroup$ Commented Mar 29, 2016 at 4:08
1
$\begingroup$

Existence of $\displaystyle \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$ means differentiability at $x$. But this question is about a different limit.

Hint: try an even function....

$\endgroup$
0
1
$\begingroup$

The may not be particularly satisfying, but put $f(x) = 0$ for $x\neq 0$ and $f(0)=1$. Then the limit exists and is zero at $x=0$ but $f$ isn't even continuous there, much less differentiable.

$\endgroup$
2
  • $\begingroup$ $f(x) = \lvert x \rvert$ works as well. $\endgroup$ Commented Mar 29, 2016 at 4:05
  • $\begingroup$ aaa i was dumb not to think about $x=0$ thanks $\endgroup$ Commented Mar 29, 2016 at 4:06
0
$\begingroup$

How about, $y = x^{2/3}$ evaluated at $0.$

Piece-wise functions are good for these sorts of problems.

$ f\left(x\right) = \left\{ \begin{array}{lr} x^2 & : x \le 0\\ x & : x > 0 \end{array} \right.\\$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.