The following equivalent statements hold:
(i) eu/u→∞e^{u}/u\rightarrow\infty as u→∞u\rightarrow\infty;
(ii) ue-u→0ue^{-u}\rightarrow 0 as u→∞u\rightarrow\infty;
(iii) xlogx→0x\log x\rightarrow 0 as x→0+x\rightarrow 0+.
Solution. (i) For u>0u>0, we have
so eu/u≥u/2e^{u}/u\geq u/2; hence eu/u→∞e^{u}/u\rightarrow\infty as u→∞u\rightarrow\infty.