Suppose $f$ is a real function defined on $R^1$ which satisfies $$\lim_{h\to 0}[f(x+h)-f(x-h)]=0$$ for every $x\in R^1$. Does this imply that $f$ is continuous?

below is my solution, my answer is yes but I looked up the solution manual which says it doesn't and I am confused which step of my reasoning is incorrect. Thank you.

Because $$\lim_{h\to 0}[f(x+h)-f(x-h)]=0$$
Let $$\lim_{h\to 0}f(x+h) = \lim_{h\to 0}f(x-h) = v = f(x)$$
Because $$\lim_{h\to 0}f(x+h) = v$$
thus $$\forall \epsilon>0, \exists h_1>0, \forall h < h_1, d(f(x+h),f(x)) < \epsilon$$
Similarly, for $f(x-h)$, I get $$\forall \epsilon>0, \exists h_2>0, \forall h < h_2, d(f(x-h),f(x)) < \epsilon$$
Let $$H = min(h_1, h_2)$$
$$\forall \epsilon>0, \exists H > 0, \forall p\in R^1, \text{if } d(x,p)<H, \text{then } d(f(x), f(p))<\epsilon$$
Thus, $f$ is continuous.

  • $\begingroup$ It is not true that $f (x)=lim_{h\to 0} f (x+h) $. $\endgroup$
    – Soma
    Commented Nov 19, 2016 at 18:35
  • $\begingroup$ See also: Does $\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0$ imply that $f$ is continuous? Other questions linked there might be of interest, too. $\endgroup$ Commented Nov 20, 2016 at 0:55
  • $\begingroup$ You should clarify whether you are asking mainly for checking your proof or finding possible errors (in which case you should use (proof-verification tag) or whether you are looking simply for any solution of this problem (in which case it is a duplicate of the question linked in the previous comment). $\endgroup$ Commented Nov 20, 2016 at 0:58
  • $\begingroup$ @MartinSleziak done $\endgroup$ Commented Nov 20, 2016 at 4:39

3 Answers 3


The mistake is in the line

Let $$\lim\limits_{h\rightarrow 0}{f(x+h)} = \lim\limits_{h\rightarrow 0}{f(x-h)} = v = f(x).$$

If $\lim\limits_{h\rightarrow 0}{f(x+h)}$ did exist, then it is true that it is equal to $\lim\limits_{h\rightarrow 0}{f(x-h)}$. However, there is no reason it should exist, and even if it did exist, there is no reason it has to equal $f(x)$.

An example of where the limit does not even exist is the function $$ f(x) = \left\{\begin{matrix} \frac{1}{x^2}& x\ne 0 \\ 0 & x = 0\end{matrix}\right.$$

  • $\begingroup$ Aptly written. +1 $\endgroup$
    – Mark Viola
    Commented Nov 19, 2016 at 19:13
  • $\begingroup$ Another example where the hypothesis of the problem is satisfied is $f(x) = \cos\frac{1}{x}$ for $x\ne 0$ and $f(0)=42$. Here again, $\lim_{h\to 0} f(0+h)$ does not exist. In ajotatxe's example (other answer) with the removable discontinuity, $\lim_{h\to 0} f(0+h)$ exists (for $h$ tending to $0$ "from both sides"), but it does not equal $f(0)$; however, we just need to change the function value in a single point to make that one continuous. $\endgroup$ Commented Nov 19, 2016 at 21:03

This is false. Take for example $f(x)=0$ for $x\neq0$ and $f(0)=1$.

The problem is in the second equation of your proof. The limit $v$ needn't be $f(x)$. In fact, $f$ is continuous if and only if the limit $v$ exists and $v=f(x)$.


From $\lim_{h\to 0}[f(x+h)-f(x-h)]=0$ it doesn't necessarily follow that $\lim_{h\to 0}f(x+h) = \lim_{h\to 0}f(x-h)$ because the two limits might not exist.


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