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:
$$x\ln{x} < x^2$$
$$\text{around zero:}-x<x\ln{x}$$
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.
 A: 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.
