Limit of this function at 0+ Let $$f(x):=  x\log\left(\frac{1}{x}\right) \exp((-\log x)^{\alpha}) .$$
where $\alpha$ is fixed. 
My question is, what is the limit $\lim_{x \downarrow 0} f(x)$ ?
This is easily shown to be $\infty$ for $\alpha \in [1,\infty)$, and also it's easy to see that this limit is $0$ for $\alpha \in (-\infty,0].$
So I am interested in the range $\alpha \in (0,1)$.
Many thanks for your help. 

Set $$y = -\log x = \log\left(\frac{1}{x}\right)$$ to get $$\begin{align}x\log(\frac{1}{x})\exp((-\log x)^\alpha)=y\exp(-y+y^\alpha) \\ = \frac{y}{\exp(y-y^{\alpha})}\end{align}$$ Then since $$\lim_{x \rightarrow 0}y= \lim_{x \rightarrow 0} (-\log(x))= \infty$$ and numerator and denominator tends to infinity you can apply L'Hospitals rule.
