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?
Best is to use the inequality $$\log t\leq t - 1,t\geq 1\tag {1}$$ Since $x\to 0^{+}$ we can assume $0&lt;x&lt;1$ so that $1/x&gt;1$ and now we have $$\log x=-2\log(1/\sqrt{x})\geq - 2(x^{-1/2}-1)&gt;-2x^{-1/2}$$ and thus we have $$|x\log x|&lt;2\sqrt{x},0&lt;x&lt;1$$ Consider any $\epsilon &gt;0$ and choose $\delta=\min(1,\epsilon ^2/4)$ then for $0&lt;x&lt;\delta$ we have $|x\log x|&lt;\epsilon$ and thus $x\log x\to 0$ as $x\to 0^{+}$.
First formulate it in the given language. So, we want to find a $\delta$ such that $|x|&lt;\delta$ implies $|x\ln(x)| &lt; \varepsilon$. Now, 

$|x\ln(x)|=|\ln(x^x)|&lt;\varepsilon$

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.