When proving a limit at $a$ with value $L$ with the definition, we must show that for all $\epsilon >0$, there is $\delta >0$ such that:

$$|x-a|<\delta \hspace{1cm} \implies \hspace{1cm} |f(x)-L|<\epsilon $$

Suppose that we come up to the following inequality when proving a certain limit:


Is the limit already proved to exist? Because given $\epsilon>0$, we have:


So, taking smaller $\epsilon$ will force both $|x-a|$ and $|f(x)-L|$ to become small so there seems there is a $\delta\leq \epsilon$, but actually finding this $\delta$ could be cumbersome.

  • Question: When proving a limit, if we find an inequality such as $(\star)$, do we need to specify $\delta$?

3 Answers 3


No. Suppose that $f=\cos$ (with domain equal to $(-1,1)$), that $L=2$ and that $a=0$. Then we always have $|x-a|<\bigl|f(x)-L\bigr|$, since$$|x-a|=|x|<1\quad\text{and}\quad\bigl|f(x)-L\bigr|=\bigl|\cos(x)-2\bigr|\geqslant1.$$However, it is not true that $\lim_{x\to0}\cos(x)=2$, and, in fact, if you take $\varepsilon=1$, no appropriate $\delta$ exists.

But if you have $\bigl|f(x)-L\bigr|<|x-a|$, then, for every $\varepsilon>0$, you can just take $\delta=\varepsilon$.


In fact $|x-a|\le|f(x)-L|$ does not imply $\lim_{x\to a}f(x)=L$. For example, let $a=L=0$ and $$f(x)=\begin{cases}1,&(|x|\le1), \\|x|,&|x|>1.\end{cases}$$

It's hard to say exactly where your error is, since I simply don't follow most of what you wrote. But you might note that the "We have...$|f(x)-L|<\epsilon$" is wrong. We don't "have" that inequality; that is, the inequality is not something we know, rather it's something we want.


To add to the other answers: the inequality $|x-a| \leq |f(x) - L|$ is a local property - this may hold for any $x$ but note that the right side still depends on $x$, so at best you can use this to show $\forall\varepsilon\forall x \exists\delta:\dots$, but you actually want to show $\forall\varepsilon\exists\delta\forall x:\dots$. You might think "let's just take the maximal (formally: the supremum) right hand side as $\delta$, surely this will also work for the other values of $x$" but this also leads to problems since this needn't exist.


You must log in to answer this question.

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