I am learning real analysis , and started with continuity of function. I am reading the following example of proving continuity of a real function with the '$\delta-\epsilon$' definition

Prove $f(x)=2x^2+1$ is continuous using the $\epsilon-\delta$ definition.

Proof : Let $x_0$ be in $\mathbb{R}$ and let $\epsilon > 0 $. We want to show that $|f(x)-f(x_0)|<\epsilon$ provided there exists a $\delta>0$, s.t $|x-x_0|<\delta$ .

We note $|f(x)-f(x_0)| = |(2x^2+1)-(2x_0^2+1)|= > |2x-2x_0^2|=2|x-x_0||x+x_0|$.We need a bound for $|x+x_0|$ that does not depend on x.. We notice that if $|x-x_0|<1$, say , then $|x|<|x_0|+1$ and hence $|x+x_0|\le |x|+|x_0|<2|x_0|+1$.

Thus we have $|f(x)-f(x_0)| \leq 2|x-x_0|(2|x_0|+1)$ provided $|x-x_0|<1$. To arrange for $2|x-x_0|(2|x_0|+1)<\epsilon$, it is enough to have $|x-x_0|<\frac{\epsilon}{2(2|x_0|+1)}$ and also $|x-x_0|<1$. So we put $\delta = min(1,\frac{\epsilon}{2(2|x_0|+1)})$. Thus we will get our desired result.

In this whole proof what I don't understand is why we need a bound for $|x-x_0|$ that does not depend on $x$. What will happen if the bound depends on $x$.

  • 1
    $\begingroup$ Ultimately your estimate has to reach $\varepsilon$, which is independent of $x$, so the dependence on $x$ has to be dropped sooner or later. $\endgroup$ Oct 21, 2019 at 7:19
  • $\begingroup$ This may be slightly against house rule, but I just wanted to say that this is an example of how to ask a good question. Up-voted. $\endgroup$
    – user284001
    Oct 21, 2019 at 8:38

2 Answers 2


Since you have to given $\epsilon$ produce (the existence of) a $\delta>0$ such that $|f(x)-f(x_0)| < \epsilon$ whenever $|x-x_0|<\delta$ (that is for any such $x$). The $|f(x)-f(x_0)|<\epsilon$ inequality must hold therefore independently of $x$ (given that $|x-x_0|<\delta$).

That is eventially you have to produce a bound for $|f(x)-f(x_0)|$ that is independent of $x$ as long as $x$ is within some bounds.

  • $\begingroup$ I think you are talking about uniformly continouous functions $\endgroup$
    – David
    Oct 21, 2019 at 7:35
  • $\begingroup$ @David No, it is allowed to depend on $x_0$. It mustn't depend on $x$ as soon as $|x-x_0|<\delta$, $|f(x)-f(x_0)|$ must be less than $\epsilon$ regardless of which such $x$ is choosen. $\endgroup$
    – skyking
    Oct 21, 2019 at 8:46
  • $\begingroup$ First we pick an $x$ and $\epsilon$, then we find $\delta$ for that particular $x$ and $\epsilon$ $\endgroup$
    – David
    Oct 21, 2019 at 11:57
  • $\begingroup$ @David No, we first pick $x_0$ and $\epsilon$ then we should be able to pick a $\delta>0$ depending on $x_0$ and $\epsilon$ such that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$ (that is for any $x$ fulfilling the last inequality) $\endgroup$
    – skyking
    Oct 21, 2019 at 12:36
  • $\begingroup$ You've just repeated exactly what I was saying. In your original answer, it could be understood that, given $\epsilon$, the same $\delta$ should work for every $x$ $\endgroup$
    – David
    Oct 21, 2019 at 13:05

To comply with the definition we need that $\delta$ exists for any $\epsilon$ such that

$$\forall x,\quad|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$$

since we are interested to the continuity around $x_0$ we can assume wlog that $x_0-1<x<x_0+1$ that is $|x-x_0|<1$ and in this case it is sufficient to assume


to satisfy the condition given by the definition.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .