# Limit of $x\ln x$ without using L'Hôpital's rule

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: $$x\ln{x} < x^2$$ $$\text{around zero:}-x The above statement was found using the following reasoning: $$\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.

• 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. Commented Apr 19, 2022 at 20:59

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 $$x\ln x . However it's not true that for all $$x\in \left]0,+\infty\right[$$ we have $$-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 $$-\sqrt{x}. Hence squeeze theorem give for all $$x\in \left]0,+\infty\right[$$ that $$-\sqrt{x} 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.
• How would you show that $-\sqrt{x}<x\ln x$ ? Commented Apr 23, 2022 at 21:16
• Define the mapping $f(x)=x\ln x+\sqrt{x}$ over $]0,+\infty[$ then see in $f'$ and use monoticity theorem. Commented Apr 23, 2022 at 21:43