real analysis - The epsilon-delta definition of continuity ...
Consider the implications of using this definition for any constant function (which should all be continuous, if any function is to be continuous).
calculus - How to prove a limit exists using the $\epsilon$-$\delta ...
So in order to make sure that $|f(x)-5|\lt \epsilon$, it is enough to require that $|x-1|\lt\epsilon$. Thus, we can select $\delta=\epsilon$. Then $\delta\gt 0$, and if $0\lt |x-1|\lt\delta$, then it will follow that $|f(x)-5|\lt\epsilon$. Thus, for all $\epsilon\gt 0$ there exists a $\delta\gt 0$ (namely, $\delta=\epsilon$) with the property ...
real analysis - Understanding $\epsilon-\delta$ definition of ...
2023年5月30日 — Yet, for any $\delta> 0, |f(0+\delta) - f(x)| \le 1$ So, $\epsilon$ (slightly greater than $1$) exists, and does not force this function to be continuous. In the traditional definition, $\epsilon$ can be arbitrarily small. Your revised definition does put that same constraint on $\epsilon.$ $\endgroup$ –
Epsilon delta theorem proof - Mathematics Stack Exchange
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Stuck on an epsilon-delta proof where I let delta be a minimum of …
Notice that we are NOT choosing $\delta =1$ or $\delta =\epsilon/3$ separately. We are choosing $\delta$ to be either $1$ or $\epsilon/3$ whichever is smaller ( We don't know which one is smaller a priori). However this choice of $\delta$ implies the above two inequalities: $\delta \leq 1$ AND $\delta \leq \epsilon/3$.
real analysis - Does the epsilon-delta definition of limits truly ...
2023年6月20日 — Epsilon delta uses the absolute value of a difference to express spatial dimension, and it appeals to the ...
How to prove that limit doesn't exist using epsilon-delta definition?
8. $\begingroup$. It is easy to prove the limit exists, all we have to show is there exists a relationship between $\delta$ and $\epsilon$. But how are we supposed to prove limit doesn't exists? The problem is when we are proving for a limit we already know what the limit is and with that, algebra is all that's needed.
calculus - How to make sense out of the $\epsilon-\delta
2015年10月21日 — For every positive $\epsilon$, there exists such a positive $\delta$ that if $|x-c|$ is smaller than $\delta$, but larger than $0$, then $|f(x) - L|$ is smaller than $\epsilon$. First of all, lets's get the double inequality out of the way.
How is the epsilon-delta definition of continuity equivalent to the ...
2015年8月16日 — Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Understanding Epsilon-Delta Proof - Mathematics Stack Exchange
2020年5月4日 — The technique is to replace the target inequality by a simpler one. Thus we have to find some expression g(x) which is simpler in form and satisfies | x2 − 9 | <g(x) and then replace the goal with ensuring that g(x) <ϵ. Thus our original target is to be achieved via a combination of two simpler goals | x2 − 9 | <g(x) and g(x) <ϵ.