Question

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

Answer 1

\(\Large\color{black}{ \displaystyle \lim_{x\rightarrow0^+ }\left( \ln x\right) }\)

Answer 2

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)

Answer 3

\[\lim(f(x)~~+g(x))~~=~~\lim(f(x))~~+\lim(g(x))\]

is valid only if both the individual limits exist

Answer 4

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....

Answer 5

(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 (-∞).

Answer 6

even then, after *applying*

Answer 7

consider a similar example :
\[\lim\limits_{x\to \infty} ~0~ = ~~\lim\limits_{x\to\infty}(\frac{1}{x}-\frac{1}{x}) \stackrel{?}{=}~\lim\limits_{x\to\infty} \frac{1}{x}~-\lim\limits_{x\to\infty} \frac{1}{x}\]

Answer 8

yeah:) lim f+lim g \(\ne\) lim (f+g) in this case.... :o

Answer 9

so there are 3 problems here. (I thought of only 1st 2 when I wrote it)

Answer 10

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

Answer 11

oh, ok. what I though is wrong was right, right, was wrong, (blah blah..

Answer 12

Would it be valid to combine the 1's or do they dissolve one-by-one?

Answer 13

I think you cannot split the limit at below step because not both the individual limits exist :