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inkyvoyd
 3 years ago
Applying the epsilondelta limit definition
inkyvoyd
 3 years ago
Applying the epsilondelta limit definition

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inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0I'll just state wikipedia's definition for my own reference. \( \forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < x  c  < \delta \ \Rightarrow \ f(x)  L < \varepsilon)\)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0the example my textbook gives is the following: "Show that the \(\lim_{x\rightarrow1}(5x3)=2\)"

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0Then, they use substitutions with what is given c=1,L=2.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0So we show this statement to be true? \(0<x1<\delta \rightarrowf(x)2<\epsilon\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes.. 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
 3 years ago
Best ResponseYou've already chosen the best response.0as \(x\to1\), \(\delta\to0\implies\epsilon\to0\)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0okay. Then the book says "We find delta by working backwards from the epsiloninequality: \((5x3)2=5x5<\epsilon\) \(x1<\epsilon/5 \)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0so they essentially manipulate the f(x)L expression to arrive at the xc expression?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that would be backward mapping... map "f(x)" to "x" given "L"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is very similar to the process of obtaining the vertical asymptote of a function.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0What I don' get is the last part. "thus we can take \(\delta=\epsilon/5\) If 0<x1<delta=epsilon/5, then \((5x3)2=5x1<5(\epsilon/5)=\epsilon\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so \(\epsilon\) exists and hence the limit is true. Also, could you tell me WHY they made that particular scalar multiple for \(\delta\)?

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0does that 5 give us any hints as to the rate the limit converges? or does it just pop up conincidentally?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ 0<x1<\delta\\ \epsilon>(5x3)2=5x1<5\delta\\ \] now, for a certain "x", \[ \epsilon=5\delta \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now, 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
 3 years ago
Best ResponseYou've already chosen the best response.0hmm 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
 3 years ago
Best ResponseYou've already chosen the best response.0in 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
 3 years ago
Best ResponseYou've already chosen the best response.0I do not understand the last question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why dont you give it a shot for f(x)=x^21 in the above process... you'd end up weith the "First principle of differentiation" :)

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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^21?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, they are not. they are "rational" functions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can play with rational numbers as long as you avoid "divisions with "0""

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0"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
 3 years ago
Best ResponseYou've already chosen the best response.0mm 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
 3 years ago
Best ResponseYou've already chosen the best response.0nice day.. started at 6am with an incredible integration problem here ending with a good talk on limits.

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0xD was it the integral of sqrt(tan x) by any chance?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dont remember exactly.. it was a form of Beta function

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0oh wow... I got a while before I'll even remotely understand that...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i htink i was\[\int_0^1\frac{\mathrm{d}x}{\sqrt{1x^m}}\]
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