*Question 1* If we have $\frac{1}{n}\sum_{i=1}^{n} \log x_{i} \leq \log m$then $\log \left[ \left( \prod_{i=1}^{n} x_{i} \right)^{\frac{1}{n}} \right] \leq \log m$hence $\left( \prod_{i=1}^{n} x_{i} \right)^{\frac{1}{n}} \leq m$since $\log$ is strictly increasing. *Question 2* Note that $0<\frac{u^{m}}{e^{u}} = \frac{u^{m}}{\sum_{i=1}^{\infty} \frac{u^{i}}{i!} } < \frac{u^{m}}{\frac{u^{m+1}}{(m+1)!}} = \frac{(m+1)!}{u} \to 0 $as $u\to \infty$. Hence, by sandwich rule, $\lim_{ u \to \infty } \frac{u^{m}}{e^{u}}=0$. *Question 3* For any $M<0$, choose $\delta=e^M$. Then if $0<x<\delta$, $\log x < \log e^{M } =M $so $\lim_{ x \to 0^{+} } \log x = -\infty$. Below is a plot for $x \mapsto x\log x$ on the interval $(0,1]$. ![[Screenshot 2024-02-05 at 02.49.51.png|300]] Take $u=-\log x \implies x=e^{-u}$ so $x\to 0^{+}$ as $u \to \infty$. Hence $0=\lim_{ u \to \infty } \frac{u}{e^{u}} = \lim_{ x \to 0^{+} } \frac{-\log x}{ \frac{1}{x}} = \lim_{ x \to 0^{+} } -x\log x $which shows that $\lim_{ x \to 0^{+} } x\log x = 0$. Note that $x^{x} =\exp(x\log x)$ so $\lim_{ x \to 0^{+} } x^{x} = \exp(0) =1 .$