This may be a stupid question but why is it that the definition of the limit is defined the way it is and not as follows $$\forall\epsilon>0\exists\delta>0\exists x(0<|x-a|<\delta\implies |f(x)-L|<\epsilon)$$

  • $\begingroup$ Because you need all $x$ in the ball to satisfy the right hand side. For example, take $a=0$ with $x \mapsto \sin {1 \over x}$. Then with $L=0$ you can find $x$ arbitrarily close to $0$ with $ \sin {1 \over x} = L$, but clearly it does not converge to $L$ (since it oscillates wildly). $\endgroup$
    – copper.hat
    Jul 12 '17 at 3:46
  • $\begingroup$ Do you mean to say that then the notion of $f(x)$ getting closer to $L$ as $x$ gets closer to $a$ will disapear $\endgroup$ Jul 12 '17 at 3:53
  • $\begingroup$ Well, with the definition above, then $L$ is not a limit in the usual sense. You need for all rather than there exists on the $x$. $\endgroup$
    – copper.hat
    Jul 12 '17 at 3:54

You want the value of just "one" point $x$ to be as close to the value of the limit for each $\epsilon$. Then the limit is not uniquely defined.

For example,let $f(x)$ on $[0,1]$ be defined as $f(x) = 1$ if $x = \frac 1n$, $n \in N$ and $0$ otherwise.

What is $\lim_{x \to 0} f(x)$? From your definition, it is both $0$ and $1$.

To see this , note that for any $\epsilon>0$, let $\delta=\epsilon$, for example, then we can find easily, some point more than $\delta$ close enough to $0$, that isn't of the form $\frac 1n$, and at this point, $f(x) = 0$, so $|f(x)-0| = 0 < \epsilon$, fitting the limit criteria.

Simiarly, given $\epsilon > 0$, let $\delta$ be smaller than $\epsilon$, and of the form $\frac 1n$. Then, if we pick $\frac \delta 2$ for example, this is more than $\delta$ close to $0$, and we have that $f(\frac \delta 2) = 1$, so that $|f(\frac \delta 2) - 1| = 0 < \epsilon$.

Hence, both $1$ and $0$ qualify as limits of the sequence under this property. However, having two different limits is an "undesirable" property, ideally when we speak of convergence in a metric space we would like unique limits. This also impedes the notion of continuity, for if there are different limits of the same sequence, how do you compare this list to a single value of a function at a point, say.

Hence, the notion of function limits you have prescribed is not suitable. However, full marks for at least thinking about an alternate approach.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.