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inkyvoyd
Applying the epsilon-delta limit definition
I'll just state wikipedia's definition for my own reference. \( \forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)\)
the example my textbook gives is the following: "Show that the \(\lim_{x\rightarrow1}(5x-3)=2\)"
Then, they use substitutions with what is given c=1,L=2.
So we show this statement to be true? \(0<|x-1|<\delta \rightarrow|f(x)-2|<\epsilon\)
yes.. the wiki definition can be elaborately expanded as below: \[ \text{if:} \lim_{x\to a^-}f(x)=\lim_{x\to a+}f(x)\\ \text{then,}\lim_{x\to a}f(x)\quad\text{exists and}\\ \lim_{x\to a^-}f(x)=\lim_{x\to a+}f(x)=\lim_{x\to a}f(x) \]
and \[\epsilon\to0\]
as \(x\to1\), \(\delta\to0\implies\epsilon\to0\)
okay. Then the book says "We find delta by working backwards from the epsilon-inequality: \(|(5x-3)-2|=|5x-5|<\epsilon\) \(|x-1|<\epsilon/5 \)
so they essentially manipulate the f(x)-L expression to arrive at the x-c expression?
that would be backward mapping... map "f(x)" to "x" given "L"
it is very similar to the process of obtaining the vertical asymptote of a function.
What I don' get is the last part. "thus we can take \(\delta=\epsilon/5\) If 0<|x-1|<delta=epsilon/5, then \(|(5x-3)-2|=5|x-1|<5(\epsilon/5)=\epsilon\)
so \(\epsilon\) exists and hence the limit is true. Also, could you tell me WHY they made that particular scalar multiple for \(\delta\)?
does that 5 give us any hints as to the rate the limit converges? or does it just pop up conincidentally?
\[ 0<|x-1|<\delta\\ \epsilon>|(5x-3)-2|=5|x-1|<5\delta\\ \] now, for a certain "x", \[ \epsilon=5\delta \]
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.
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.
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)
I do not understand the last question
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" :)
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?
no, they are not. they are "rational" functions
you can play with rational numbers as long as you avoid "divisions with "0""
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...
"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....
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!
nice day.. started at 6am with an incredible integration problem here ending with a good talk on limits.
xD was it the integral of sqrt(tan x) by any chance?
dont remember exactly.. it was a form of Beta function
oh wow... I got a while before I'll even remotely understand that...
i htink i was\[\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^m}}\]