Clarification on limit definition.

Question

I know that in the definition of limit they say that if $\ \ \forall \varepsilon > 0\ \ \exists \ \delta > 0 \ \ $But why not $\ \ \forall \delta> 0\ \exists \ \varepsilon > 0$?

Thanks for the replies. But, my question is not just about inter changing greek letters. I am asking, why the delta on the x axis is dependent on the epsilon on the y axis? why not the other way around?

Answer 1

What matters is not the Greek letter that's used after the universal quantifier and the existential quantifiers. What matters is the rest of the definition, and matching the correct variable in the ensuing definition with the correct quantified variable.

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$$\forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)$$ could just as easily be written (albeit unconventionally) as

$$\forall \delta > 0\ \exists \ \varepsilon > 0 : \forall x\ (0 < |x - c | < \varepsilon \ \Rightarrow \ |f(x) - L| < \delta). $$

Answer 2

We will assume that you are proposing to define $\lim_{x\to a}f(x)=b$ as follows. We say that $\lim_{x\to a} f(x)=b$ if for every $\delta\gt 0$ there is an $\epsilon\gt 0$ such that if $0\lt |x-a|\lt \delta$ then $|f(x)-b|\lt \epsilon$. In other words, we are assuming that the rest of the definition is word for word the standard one.

Let $f(x)=\sin(1/x)$ for $x\ne 0$. A picture will show that $\lim_{x\to 0} f(x)$ does not exist. One can even do it without a picture. For when $x=\frac{1}{2n\pi}$, then $\sin(1/x)=0$, while if $x=\frac{1}{(2n+1/2)\pi}$, then $\sin(1/x)=1$. In fact near $0$, the function $\sin(1/x)$ bounces wildly between $-1$ and $1$.

Let $b=3$. We show that under your proposed definition, $\lim_{x\to 0} f(x)=b$. Pick any positive $\delta$, and let $\epsilon=17$. Then when $0\lt |x-0|\lt \delta$, we have $|f(x)-3|\lt \epsilon$.

There is no need to pick a "wild" function. Let $f(x)=\sin x$. The same reasoning, under the proposed definition, will show that $\lim_{x\to 0} f(x)=3$. For to every $\delta\gt 0$, there is an $\epsilon$, namely $17$, such that if $|x-0|\lt \delta$, then $|\sin x-3|\lt 17$.