# When proving a limit, does finding $|x-a|\leq|f(x)-L|$ suffices to show that there is a $\delta$?

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:

$$|x-a|\leq|f(x)-L|\tag{\star}$$

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

$$|x-a|\leq|f(x)-L|<\epsilon$$

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 $$\delta$$?

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.