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July 5th, 2019, 12:31 PM   #1
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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.
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July 6th, 2019, 04:19 AM   #2
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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 $|f(x)-2| < \epsilon$.
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July 6th, 2019, 03:08 PM   #3
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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.
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