I know that there are already numerous questions on $\lim\limits_{x\to0}x\ln(x)$. However, I am interested in a specific trait when it comes to prove the existence of the limit by the use of rule of L'Hopital.
Why is it a dead end if we start off like this:
$$
\lim\limits_{x\to0}x\ln(x)=\frac{\lim\limits_{x\to0}x}{\lim\limits_{x\to0}\frac{1}{\ln(x)}}\underset{\text{L'Hopital}}{=}\frac{\lim\limits_{x\to0}1}{\lim\limits_{x\to0}\frac{(-1)}{x\ln(x)^2}}=\lim\limits_{x\to0}(-x)\ln(x)^2=\cdots?
$$
If we start with the right expression then we see that the limit exists and if not we get lost. This seems a bit unsatisfying as it involves a kind of "luck". Maybe there is a deeper sense in this point which I don't see. Or did I simply made a mistake?
Hope this question is not too vague. If it is then let me know and I will delete it.
Best Answer
The equality$$\lim_{x\to0}x\log(x)=\frac{\lim_{x\to0}x}{\lim_{x\to0}\frac1{\log x}}$$is false, since the LHS is $0$, whereas the RHS doesn't make sense. And the equality$$\frac{\lim_{x\to0}x}{\lim_{x\to0}\frac1{\log x}}=\frac{\lim_{x\to0}1}{\lim_{x\to0}-\frac1{x\log^2(x)}}$$simply says that a thing that doesn't make sense is equal to another thing which also doesn't make sense.
On the other hand, it is indeed true that if you try to apply L'Hopital's rule to compute the limit$$\lim_{x\to0}\frac{x}{\frac1{\log x}}$$you get another indeterminate forme. So what? That happens a lot.