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I was trying to find

\begin{equation} \lim_{x \to 0^{+}}{x\ln{x}}. \end{equation}

Note that I let $x$ approach from right hand side only because $\ln{x}$ is undefined within $x < 0$ in $\mathbb{R}$.

Now, because the limit gives '$0\cdot-\infty$', I need to convert it into form '$\frac{0}{0}$' to use l'Hôpital's rule. Therefore, I have

\begin{equation} \lim_{x \to 0^{+}}{\frac{x}{\frac{1}{\ln{x}}}}. \end{equation}

Then, by applying l'Hôpital's rule, I obtain

\begin{equation} \lim_{x \to 0^{+}}{\frac{x}{\frac{1}{\ln{x}}}} = \lim_{x \to 0^{+}}{\frac{1}{-\frac{1}{(\ln{x})^2}\frac{1}{x}}} \iff \lim_{x \to 0^{+}}{x\ln{x}} = -\lim_{x\to0^{+}}{\ln{x}} \cdot \lim_{x\to0^{+}}{x\ln{x}} \end{equation}

In the equation above, I observe that there is $\lim_{x \to 0^{+}}{x\ln{x}}$ on both sides. If I divide both sides by $\lim_{x \to 0^{+}}{x\ln{x}}$, I obtain

\begin{equation} \lim_{x\to0^{+}}{\ln{x}} = -1, \end{equation}

which is absolutely nonsense because I already know that $\lim_{x\to0^{+}}{\ln{x}} = -\infty$.

So, the division above is not possible. The only reason that I have come up with is that the only way that a division is not allowed is that the divisor is $0$. Therefore, there must be $\lim_{x \to 0^{+}}{x\ln{x}} = 0$.

Therefore, by the argument above, I have shown that $\lim_{x \to 0^{+}}{x\ln{x}} = 0$. I am wondering is this a valid argument? Is there anything incorrect in this argument? Do you agree with me? I want to here your opinion!

EDIT: I understand that I can use $\frac{\ln{x}}{\frac{1}{x}}$ for the l'Hôpital's rule because it is of the form '$\frac{\infty}{\infty}$'. However, I also know that, unlike the form '$\frac{0}{0}$', l'Hôpital's rule doesn't always hold for the form the form '$\frac{\infty}{\infty}$'. Therefore, I chose to use $\frac{x}{\frac{1}{\ln{x}}}$ instead.

Bruce M
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    There are several errors in your argument. For a starter, $\lim _{x\to 0^+}{\frac {f(x)}{g(x)}}=\lim _{x\to 0^+}{\frac {f'(x)}{g'(x)}}$ holds only if $\lim _{x\to 0^+}{\frac {f'(x)}{g'(x)}}$ is known to exist. – Martin R Dec 17 '23 at 10:14
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    Things become much simpler if you write $x \ln x = \frac{\ln x}{1/x}$, see for example https://math.stackexchange.com/q/1052932/42969 or https://math.stackexchange.com/q/470952/42969. – Martin R Dec 17 '23 at 10:17
  • The same error Martin R observed happens when you change $\lim_{x \to 0^{+}}{\frac{1}{-\frac{1}{(\ln{x})^2}\frac{1}{x}}}$ to $-\lim_{x\to0^{+}}{\ln{x}} \cdot \lim_{x\to0^{+}}{x\ln{x}}$. The theorem about limits is: The limit $\lim (f\cdot g)$ exists and equals $(\lim f)\cdot(\lim g) \color{red}{\text{ as long as the limits } \lim f \text{ and } \lim g \text{ exists}}$. – jjagmath Dec 17 '23 at 11:41
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    Does this answer your question? Limit of $x \log x$ as $x$ tends to $0^+$ – Jam Dec 17 '23 at 11:43
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    By substitution $x=e^{-t}$ the limit is equal $-\lim_{t\to \infty}{t\over e^t}.$ – Ryszard Szwarc Dec 17 '23 at 16:22
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    You divide by $\lim x\ln x$. You claim you cannot do this, and the only way for this to be invalid is when the limit is zero. But in fact it is also impossible and invalid reasoning when the (finite) limit does not exist to begin with. – freakish Dec 17 '23 at 19:48

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