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SolomonZelman
 one year ago
errors/fallacy/valid/correct .... BUT
consider this limit (in a q.) (it is rather a joke)
SolomonZelman
 one year ago
errors/fallacy/valid/correct .... BUT consider this limit (in a q.) (it is rather a joke)

This Question is Closed

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right) }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I am sure (most of) you are able to demonstrate how this limit diverges to ∞, using basic algebraic and logarithmic properties. ` W A T C H M E ! ` \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \frac{x~\ln x}{x}\right) }\) since top and bottom are 0 when you plug in 0 for x, we are able to use LHS. \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \frac{\ln x+(x/x)}{1}\right) }\) \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x+1\right) }\) \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right)+\lim_{x\rightarrow0^+ }\left( 1\right) }\) \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right)+1 }\) So far, all I did is that I made this limit +1, but as we know ((∞)+1=∞) Thus, I have NOT (yet) changed the value of the limit. Now, repeat the same thing with the limit, you will have \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right)+(1\times \infty ) }\) For those that would like to write this properly, \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right)+\lim_{k\rightarrow \infty}(1\times k ) }\) then perhaps, \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right)+\lim_{k\rightarrow 0^+}(\frac{1}{k} ) }\) would the following be invalid? \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right)+\lim_{x\rightarrow 0^+}(\frac{1}{x} ) }\) \(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x+ \frac{1}{x}\right) }\) \(\Large\color{black}{ \displaystyle =\infty }\) (that limit was → ∞ previously)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim(f(x)~~+g(x))~~=~~\lim(f(x))~~+\lim(g(x))\] is valid only if both the individual limits exist

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1FIne, that is one piece.... but even then, a limit turned out to be "indeterminate" then, and I didn't seem to use anything invalid previously....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1(1) The way I see it, is that I multiplied top and bottom by 0/0 (essentially). (2) But even then after appalling LHS multiple times, I don't get to combine the one's. Each time I have a +1, it should dissolve in the result that the limit  in (∞).

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1even then, after *applying*

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1consider a similar example : \[\lim\limits_{x\to \infty} ~0~ = ~~\lim\limits_{x\to\infty}(\frac{1}{x}\frac{1}{x}) \stackrel{?}{=}~\lim\limits_{x\to\infty} \frac{1}{x}~\lim\limits_{x\to\infty} \frac{1}{x}\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yeah:) lim f+lim g \(\ne\) lim (f+g) in this case.... :o

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so there are 3 problems here. (I thought of only 1st 2 when I wrote it)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1there is nothing wrong with multiplying/dividing by \(x\) as \(x\to 0^+\) notice that \(x\) is not identically equal to 0, so it is fine to multiply/divide

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, ok. what I though is wrong was right, right, was wrong, (blah blah..

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Would it be valid to combine the 1's or do they dissolve onebyone?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think you cannot split the limit at below step because not both the individual limits exist : dw:1435585129527:dw
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