SolomonZelman
  • SolomonZelman
errors/fallacy/valid/correct .... BUT consider this limit (in a q.) (it is rather a joke)
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
\(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right) }\)
SolomonZelman
  • SolomonZelman
I 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
  • ganeshie8
\[\lim(f(x)~~+g(x))~~=~~\lim(f(x))~~+\lim(g(x))\] is valid only if both the individual limits exist

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SolomonZelman
  • SolomonZelman
FIne, 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
  • SolomonZelman
(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
  • SolomonZelman
even then, after *applying*
ganeshie8
  • ganeshie8
consider 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
  • SolomonZelman
yeah:) lim f+lim g \(\ne\) lim (f+g) in this case.... :o
SolomonZelman
  • SolomonZelman
so there are 3 problems here. (I thought of only 1st 2 when I wrote it)
ganeshie8
  • ganeshie8
there 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
  • SolomonZelman
oh, ok. what I though is wrong was right, right, was wrong, (blah blah..
SolomonZelman
  • SolomonZelman
Would it be valid to combine the 1's or do they dissolve one-by-one?
ganeshie8
  • ganeshie8
I think you cannot split the limit at below step because not both the individual limits exist : |dw:1435585129527:dw|

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