let $f\colon \Bbb R \to \Bbb R$ be a function having a property as given $$ \lim_{h \to 0} {f(x+h)-f(x-h)\over h}$$ exists for all $x \in \Bbb R$ then $f(x)$ is differentiable for all $x \in \Bbb R$. True/false??

I think its true as $$ \lim_{h\to 0} {f(x+h)-f(x-h)\over h} ...(1)$$ exists for all $x \in \Bbb R$ so put $x=x+h$ in $(1)$ so $$ \lim_{h\to 0} {f(x+2h)-f(x)\over h}$$ exists i.e $$ 2\lim_{h\to 0} {f(x+2h)-f(x)\over 2h}$$ exists and taking $k=2h$ then $$ 2\lim_{k\to 0} {f(x+k)-f(x)\over k}$$ exists i.e $2f^{'}(x)$ exists so $f^{'}(x)$ exists for all $x \in \Bbb R$.
Is this is right??


Think of $f(x) = \begin{cases}0 \text{ if } x \ne 0\\1 \text{ if } x = 0\end{cases}$. It obeys your property (easy to prove) but it's not even continuous in $0$.

The point is that with your property you're sampling your function only outside of the point where you're computing the limit, therefore the limit value may not match the actual function value at that point.

In you reasoning, replacing $x+h$ by $x$ is illegal, because you're changing a term that is variable in $h$ with a term independent of $h$.


Consider $$f(x) = \begin{cases}x \text{ if } x \geq 0 \\ -x \text{ if } x < 0 \end{cases}.$$ Then at $x=0$ $$\lim_{h \to 0} \frac{f(h)-f(-h)}{h} = \lim_{h \to 0} \frac{h-h}{h}=0.$$ But $f$ is not differentiable at $x=0$. In the rest of its domain $f$ is trivially smooth.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.