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

Oct 2015
141
8
Greece
\(\displaystyle \lim_{x \to c} f(x) = L\) 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:
1.jpg

Exercise 1 (continues):
2.jpg

Exercise 2:
3.jpg

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.
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
The limit definition says that 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 $|f(x)-2| < \epsilon$.
 
  • Like
Reactions: 2 people

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
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.
 
  • Like
Reactions: 1 person