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

1. SolomonZelman

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

2. 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)

3. ganeshie8

$\lim(f(x)~~+g(x))~~=~~\lim(f(x))~~+\lim(g(x))$ is valid only if both the individual limits exist

4. 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....

5. 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 (-∞).

6. SolomonZelman

even then, after *applying*

7. ganeshie8

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

8. SolomonZelman

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

9. SolomonZelman

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

10. 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

11. SolomonZelman

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

12. SolomonZelman

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

13. ganeshie8

I think you cannot split the limit at below step because not both the individual limits exist : |dw:1435585129527:dw|