How to calculate the limit $\displaystyle\lim_{x \to 0} x\ln(x)$ by $\varepsilon-\delta$ definition? I know limit of $x \ln(x) $ at $ 0$ is $0$ (by writing this as  $\ln(x) / (1/x)$ then using L'hospital rule) but i want to prove this by $\epsilon-\delta$ definition. Can any one give some hint?
 A: Best is to use the inequality $$\log t\leq t - 1,t\geq 1\tag {1}$$ Since $x\to 0^{+}$ we can assume $0<x<1$ so that $1/x>1$ and now we have $$\log x=-2\log(1/\sqrt{x})\geq - 2(x^{-1/2}-1)>-2x^{-1/2}$$ and thus we have $$|x\log x|<2\sqrt{x},0<x<1$$ Consider any $\epsilon >0$ and choose $\delta=\min(1,\epsilon ^2/4)$ then for $0<x<\delta$ we have $|x\log x|<\epsilon$ and thus $x\log x\to 0$ as $x\to 0^{+}$.
A: First formulate it in the given language. So, we want to find a $\delta$ such that $|x|<\delta$ implies $|x\ln(x)| < \varepsilon$. Now, 
$|x\ln(x)|=|\ln(x^x)|<\varepsilon$
for example. These kinds of manipulatons and tricks (multiplying by 1 or adding 0 in a clever way) are what will get you there. 
Note that some limits are very amenable to $\varepsilon-\delta$ proofs directly which is we have those nice tricks. 
A: Let $f(x)= x\ln x$ for $x>0.$ For $0<x<1/e$ we have $f(x)<0$ and $f'(x)<0.$ So $L=\lim_{x\to 0^+}f(x)$ exists and is not positive.
Suppose $L<0.$  Then for all sufficiently large $n\in \Bbb N$ we  have $$(*)\quad 0>L/2>e^{-n}\ln e^{-n}=-\frac {n}{e^n}.$$ But for $2\leq n\in \Bbb N,$ by the Binomial Theorem we have $e^n>2^n=(1+1)^n=1+\binom {n}{1}+\binom {n}{2}+....>\binom {n}{2}=\frac {n^2-n}{2},$ which will contradict $(*)$ if $n$ is large enough.
