My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

Thanks Tree3Thanks
  • 2 Post By v8archie
  • 1 Post By v8archie
LinkBack Thread Tools Display Modes
July 5th, 2019, 12:31 PM   #1
Senior Member
Joined: Oct 2015
From: Greece

Posts: 137
Thanks: 8

I can't understand how to prove limits using the definition...

$\displaystyle \lim_{x \to c} f(x) = L$ if $\displaystyle \forall ε > 0$, $\displaystyle \exists δ>0$ :

$\displaystyle 0 < |x-c| < δ$ => $\displaystyle |f(x) - L| < ε$ , $\displaystyle \forall x$ maybe except c.

First of all do I understand what the definitions says? My understanding is this:
"The closer x is at c the more precisely f(x) will approach L." In order to define this mathematically we need two things. An number δ to precisely specify the words "The closer x is at c" and another number ε which precisely describes the words "the more precisely f(x) will approach L". If ε is really small, the more precise the limit will be.

Some exercises which I'm trying to solve:
Exercise 1:

Exercise 1 (continues):

Exercise 2:

For example in exercise 1 he finds out that: $\displaystyle |x-1| < ε/5$
Then i can not see why he chooses δ=ε/5 . Is it because both in-equations are similar?

$\displaystyle |x-1| < ε/5$
$\displaystyle 0 < |x-1| < δ$

so it seems like δ = ε/5 if you compare them. Is there a more mathematically procedure to actually show that?

Can you explain the solution of ex1 better that the book?

Thank you.
babaliaris is offline  
July 6th, 2019, 04:19 AM   #2
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,671
Thanks: 2651

Math Focus: Mainly analysis and algebra
The limit definition says that [B]given a value for $\epsilon$, we can always pick a value for $\delta$ (the maximum difference from the limit point) such that the function is within $\epsilon$ of the limit.

So, given any $\epsilon$ we find that $|x-1| < \frac{\epsilon}{5}$. So whatever $\epsilon$ may be, if we pick our $\delta$ to be $\frac{\epsilon}{5}$, we will have $|x-1| < \frac{\epsilon}{5} = \delta$ which, since $a < b = c \implies a < c$ gives us the $|x-1| < \delta$ which we require.

All of that is preparatory work. What follows is the proof that picking $\delta = \frac{\epsilon}{5}$ delivers the required inequality $|f(x)-2| < \epsilon$.
Thanks from romsek and babaliaris
v8archie is offline  
July 6th, 2019, 03:08 PM   #3
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,671
Thanks: 2651

Math Focus: Mainly analysis and algebra
It's worth pointing out that, although we found (in the first part) an inequality that suggested a value for $\delta$, the method doesn't always indicate that there are no other places where $f(x)$ is close to $L$. This is why we need to do the second part, which is the actual proof.

Also, since $|x-1| < \frac{\epsilon}{5}$ describes a symmetrical interval about $x=1$ (for any given value of $\epsilon$), it should be clear that we could instead pick a smaller value for $\delta$ if we so desired, $\frac{\epsilon}{10} $ for example.
Thanks from babaliaris
v8archie is offline  

  My Math Forum > College Math Forum > Calculus

definition, limits, prove, understand

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
I don't understand limits and their relationship to horizontal asymptotes. sgtsloth25 Calculus 16 April 7th, 2015 12:33 PM
Use the ?-? definition of a limit to prove this. danx Calculus 5 September 25th, 2013 12:35 PM
limits epsilon delta definition dtn_perera Calculus 2 August 9th, 2013 06:46 AM
Prove C-K equation using markov chain definition Franny Algebra 0 March 29th, 2010 12:54 PM
Prove the definition of e 450081592 Calculus 1 February 5th, 2010 11:36 PM

Copyright © 2019 My Math Forum. All rights reserved.