Applying the epsilon-delta limit definition

- inkyvoyd

Applying the epsilon-delta limit definition

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- inkyvoyd

I'll just state wikipedia's definition for my own reference.
\( \forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)\)

- anonymous

ok

- inkyvoyd

the example my textbook gives is the following:
"Show that the \(\lim_{x\rightarrow1}(5x-3)=2\)"

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- inkyvoyd

Then, they use substitutions with what is given
c=1,L=2.

- inkyvoyd

So we show this statement to be true? \(0<|x-1|<\delta \rightarrow|f(x)-2|<\epsilon\)

- anonymous

yes..
the wiki definition can be elaborately expanded as below:
\[
\text{if:} \lim_{x\to a^-}f(x)=\lim_{x\to a+}f(x)\\
\text{then,}\lim_{x\to a}f(x)\quad\text{exists and}\\
\lim_{x\to a^-}f(x)=\lim_{x\to a+}f(x)=\lim_{x\to a}f(x)
\]

- anonymous

and \[\epsilon\to0\]

- anonymous

as \(x\to1\), \(\delta\to0\implies\epsilon\to0\)

- inkyvoyd

okay. Then the book says "We find delta by working backwards from the epsilon-inequality:
\(|(5x-3)-2|=|5x-5|<\epsilon\)
\(|x-1|<\epsilon/5 \)

- inkyvoyd

so they essentially manipulate the f(x)-L expression to arrive at the x-c expression?

- anonymous

that would be backward mapping... map "f(x)" to "x" given "L"

- anonymous

called image

- anonymous

it is very similar to the process of obtaining the vertical asymptote of a function.

- inkyvoyd

What I don' get is the last part.
"thus we can take \(\delta=\epsilon/5\) If 0<|x-1|

- anonymous

so \(\epsilon\) exists and hence the limit is true.
Also, could you tell me WHY they made that particular scalar multiple for \(\delta\)?

- anonymous

ooooh

- inkyvoyd

does that 5 give us any hints as to the rate the limit converges? or does it just pop up conincidentally?

- anonymous

\[
0<|x-1|<\delta\\
\epsilon>|(5x-3)-2|=5|x-1|<5\delta\\
\]
now, for a certain "x",
\[
\epsilon=5\delta
\]

- anonymous

now, what you said about hint of "derivatives" is not wrong..
Leibnitz's approach in declaring \[\lim_{\delta\to0}{\epsilon\over\delta}={dy\over dx}\]
was not a fluke...
many stalwarts did not realize that connection till the former dared to state it.

- inkyvoyd

hmm- this is a bit offtopic, but why is it that one can often treat a differential (and differential notation) as actual numbers, but they don't follow all rules? I heard somewhere that one can redefine differentials as another set of numbers, but there are rules that they dont' follow that real numbers do follow.

- anonymous

in fact, if you search into the "Antique stronghold" of "Project Gutenberg" and "Google Books", you will stumble into the rare Math bibles.
(I was very upset on the concept of eigen values and I found the history of development of matrices on Gutenberg)

- anonymous

I do not understand the last question

- anonymous

why dont you give it a shot for f(x)=x^2-1 in the above process...
you'd end up weith the "First principle of differentiation" :)

- inkyvoyd

well, leibniz interpreted dy/dx as an infinitisimal change of y divide by an infinitismal change in x. You can do things with differential form like solving separable differential equations such as dy/dx=1/x
you rewrite in differential form to get dy=1/x dx or y=ln x (differential form rewriting is like treating the notation as a fraction?)
also, you could just directly integrate
Integral dy/dx dx=integral 1/x dx
integral dy=ln x
y=ln x
The notation is weird because it suggests that dy and dx are atual numbers, but everyone says they aren't.
what should I do with f(x)=x^2-1?

- anonymous

no, they are not. they are "rational" functions

- anonymous

you can play with rational numbers as long as you avoid "divisions with "0""

- inkyvoyd

yes, but doesn't dx approach zero? I mean, youg et the indeterminant form 0/0 in all derivatives when evalating the difference quotient limit with direct substitution...

- anonymous

"approach", yes, "equal it?" no
you will still have some infinitesimally small \(\delta\), but never "0"
to avoid these problems, you use L'Hopital's Rule....

- inkyvoyd

mm- I'ma go ahead and try some practice with this epsilon delta stuff- my main motivation is to use it for riemann sums, but I'll have to understand this first. Thanks for all the help!

- anonymous

surething

- anonymous

nice day.. started at 6am with an incredible integration problem here
ending with a good talk on limits.

- inkyvoyd

xD was it the integral of sqrt(tan x) by any chance?

- anonymous

dont remember exactly.. it was a form of Beta function

- inkyvoyd

oh wow... I got a while before I'll even remotely understand that...

- anonymous

i htink i was\[\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^m}}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.