I came across the question of solving $\lim_{x\to 0^+} x\ln{x}$ by using the squeeze/sandwich theorem. Furthermore, any use of L'Hôpital's rule is not permitted. My working can be found below, yet I am not sure it can be considered correct: $$x\ln{x} < x^2$$ $$\text{around zero:}-x<x\ln{x}$$ The above statement was found using the following reasoning: $$\lim_{x\to 0^+} \frac{-x}{x\ln{x}}=\lim_{x\to 0^+} \frac{-1}{\ln{x}} =0$$ This should mean the denominator is larger than the numerator around zero. Is this reasoning correct? If it is correct, using the squeeze theorem, the limit of $x\ln{x}$ as x approaches zero from the positive direction would be zero.

  • $\begingroup$ You are correct that $x \ln(x) < x^2$ and $x^2\to0$ as $x\to0^+$, but you also need a lower bound in order to apply the squeeze theorem. $-\sqrt{x}$ seems to be reasonable choice. $\endgroup$
    – user170231
    Commented Apr 19, 2022 at 20:59

1 Answer 1


Define the mapping $x\mapsto x\ln x$ over $\left]0,+\infty\right[$. So it's true that for all $x\in \left]0,+\infty\right[$ we have $x\ln x <x^{2}$. However it's not true that for all $x\in \left]0,+\infty\right[$ we have $-x<x\ln x$, in fact it's only true when $x\in \left]e^{-1},+\infty\right[$. So your lower bound doesn't work. Now, you can use the hint given by user170231 because for all $x\in \left]0,+\infty\right[$ we have $-\sqrt{x}<x\ln x$. Hence squeeze theorem give for all $x\in \left]0,+\infty\right[$ that $-\sqrt{x}<x\ln x<x^{2}$ so $\displaystyle \lim_{x\to 0^{+}}-\sqrt{x}<\lim_{x\to 0^{+}}x\ln x<\lim_{x\to 0^{+}}x^{2}$ implies $\displaystyle \lim_{x\to 0^{+}}x\ln x=0$ so done.

There are all kinds of arguments for this problem another approach is using the inequality $(*)$ for all $x\in \Bbb{R}$, we know $e^{x}\geqslant 1+x$ . So via substitution $x\mapsto \frac{1}{e^x}$ we have $\displaystyle \lim_{x\to 0^{+}}x\ln x=\lim_{x\to +\infty}\frac{-x}{e^x}$. By $(*)$ we have for all $x\in \left]0,+\infty\right[$ that $e^{x}\geqslant (1+x/2)(1+x/2)>x^{2}/4$. Then $\displaystyle \lim_{x\to +\infty}\left|\frac{-x}{e^{x}}\right|<\lim_{x\to +\infty} \left|\frac{4}{x}\right|$ so $\displaystyle \lim_{x\to +\infty}\frac{-x}{e^x}=0$ hence again $\displaystyle \lim_{x\to 0^{+}}x\ln x=0$ by squeeze theorem.

  • $\begingroup$ How would you show that $-\sqrt{x}<x\ln x$ ? $\endgroup$
    – Cantor
    Commented Apr 23, 2022 at 21:16
  • $\begingroup$ Define the mapping $f(x)=x\ln x+\sqrt{x}$ over $]0,+\infty[$ then see in $f'$ and use monoticity theorem. $\endgroup$
    – A. P.
    Commented Apr 23, 2022 at 21:43

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