1

I am using Thomas/Finney's Calculus and Analytical Geometry and it says "Given any positive radius ε about L, there exists some positive radius δ about c such that for all t within δ units of c (except t=c itself) the values of F(t) lies within ε units of L." Here, F(t) = (f(x+t)-f(x))/(t) Where t is the single variable and we were supposed to(which I did above) set about defining what it means for F to have a limit L as t approaches a predetermined value c.

Jignesh Shah
  • 55

3 Answers3

1

The above thing is say if $\lim_{t \to c} F(t) = L$ if and only if for all $\epsilon > 0$, $\exists \delta > 0$ such that for all $t$, if $0<|t-c|<\delta$ then $|F(t) - L | < \epsilon$


Intuitively this is saying we can make the distance between $F(t)$ and $L$ arbitrarily small, by making $t$ close enough to $c$, $i.e.$, exhibiting the appropriate $\delta$.

Here's an easy example. Take $F(t) = 3t$. Say you want to prove $$\lim_{t\to 3} F(t) = 9$$,

This means that for an arbitrarily small real number $\epsilon > 0$, you need to find find a a positive real number $\delta$ so that if $t$ is within distance $\delta$ of $3$, the absolute difference, (distance) between $F(t)$ and $9$ is less than this small number $\epsilon$.

Here's where you can do some scratchwork:

The inequality: $|F(t) - 9| < \epsilon$ is: $$ |3t-9|< \epsilon \leftrightarrow |3|\cdot |t-3| < \epsilon \leftrightarrow |t-3| < \epsilon/3 $$ So no matter what positive real number $\epsilon$ you have, if you make the distance between $t$ and $3$ less than $\delta = \epsilon/3$, i.e. if $0<|t-3|< \epsilon/3 = \delta $ then you'll have: $3|t-3| = |3t-9| = |F(t) - 9| < 3\cdot \epsilon/3 = \epsilon$.

So the limit is proved.

Rustyn
  • 8,407
  • Thats the same thing written in book. – Jignesh Shah Sep 13 '14 at 13:12
  • Yeah, what's your question about it? Do you want an example? – Rustyn Sep 13 '14 at 13:13
  • I just need to know what 'ε' and 'δ' is. – Jignesh Shah Sep 13 '14 at 13:15
  • He couldn't understand what was written in the book. I am too having trouble proving limit exist using Delta Epsilon definition of limit. It is so confusing. – Neer Sep 13 '14 at 13:16
  • I posted an example. – Rustyn Sep 13 '14 at 13:23
  • The $\epsilon$'s are the positive real numbers that you're hoping the distance between $F(t)$ and $L$ can be made to be within by exhibiting the $\delta$'s which are the positive real numbers that tell us how close $t$ needs to be to $c$ in order to ensure this. ($F(t)$ and $L$ will be within $\epsilon$ if and only if we can find this $\delta$) – Rustyn Sep 13 '14 at 13:35
1

The Delta-Epsilon definition basically states that for every point ( however close that may be to the limiting point but not exactly on that point) there would a be point in y-axis near the limiting value, for a limit to exist.

enter image description here

Neer
  • 674
  • 3
  • 9
  • 22
0

You can translate that $\epsilon, \delta$ definition of the limit of a function, to sequences, note that these definition are equivalent.

The limit of a function $f(x)$ at point $a$ is $L$, if for every sequence $a_n$ where $$\lim_{n \to \infty}a_n = a$$ It follows that $$ \lim_{n \to \infty}f(a_n) = L $$

Note that members of the sequences $a_n$ must never be equal to $a$, because the function $f(x)$ might not even be defined at $a$, and even if it is defined, the value $f(a)$ makes no difference to the limit at $a$.

When you unwrap these limits of sequences to their corresponding definitions in terms of $\epsilon$, you get the $\epsilon, \delta$ definition of the limit of the function that you are having trouble with.

vladimirm
  • 1,028