inkyvoyd
  • inkyvoyd
Applying the epsilon-delta limit definition
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
inkyvoyd
  • 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
  • anonymous
ok
inkyvoyd
  • 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
  • inkyvoyd
Then, they use substitutions with what is given c=1,L=2.
inkyvoyd
  • inkyvoyd
So we show this statement to be true? \(0<|x-1|<\delta \rightarrow|f(x)-2|<\epsilon\)
anonymous
  • 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
  • anonymous
and \[\epsilon\to0\]
anonymous
  • anonymous
as \(x\to1\), \(\delta\to0\implies\epsilon\to0\)
inkyvoyd
  • 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
  • inkyvoyd
so they essentially manipulate the f(x)-L expression to arrive at the x-c expression?
anonymous
  • anonymous
that would be backward mapping... map "f(x)" to "x" given "L"
anonymous
  • anonymous
called image
anonymous
  • anonymous
it is very similar to the process of obtaining the vertical asymptote of a function.
inkyvoyd
  • inkyvoyd
What I don' get is the last part. "thus we can take \(\delta=\epsilon/5\) If 0<|x-1|
anonymous
  • 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
  • anonymous
ooooh
inkyvoyd
  • inkyvoyd
does that 5 give us any hints as to the rate the limit converges? or does it just pop up conincidentally?
anonymous
  • anonymous
\[ 0<|x-1|<\delta\\ \epsilon>|(5x-3)-2|=5|x-1|<5\delta\\ \] now, for a certain "x", \[ \epsilon=5\delta \]
anonymous
  • 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
  • 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
  • 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
  • anonymous
I do not understand the last question
anonymous
  • 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
  • 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
  • anonymous
no, they are not. they are "rational" functions
anonymous
  • anonymous
you can play with rational numbers as long as you avoid "divisions with "0""
inkyvoyd
  • 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
  • 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
  • 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
  • anonymous
surething
anonymous
  • anonymous
nice day.. started at 6am with an incredible integration problem here ending with a good talk on limits.
inkyvoyd
  • inkyvoyd
xD was it the integral of sqrt(tan x) by any chance?
anonymous
  • anonymous
dont remember exactly.. it was a form of Beta function
inkyvoyd
  • inkyvoyd
oh wow... I got a while before I'll even remotely understand that...
anonymous
  • 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.