*Question 1* Note that $f'(t)= \frac{1}{1+t} +\frac{1}{1-t}-2=\frac{2}{1-t^{2}} -2 \geq 2-2=0$ since $|t|<1\implies 0< 1-t^{2}\leq 1$. Now $f(0)=\log(1)=0$ so for $0\leq t<1$, $f(t)\geq 0$ since $f$ is increasing. Hence $\log\left( \frac{1+t}{1-t} \right) \geq 2t \tag{1}$for $0\leq t<1$. For $x>0$, let $t=\frac{1}{2x+1}$ then $2x+1>1 \implies 0<t<1$. Also $\frac{1+t}{1-t}=\frac{ 1+\frac{1}{2x+1} }{1- \frac{1}{2x+1}} = \frac{2x+2}{2x} = 1+\frac{1}{x}$Substituting in $(1)$ gives $\log\left( 1 + \frac{1}{x} \right) \geq \frac{2}{2x+1}$Rearranging gives $\left( x+\frac{1}{2} \right)\log\left( 1+\frac{1}{x} \right) \geq 1 \implies \left( 1+\frac{1}{x} \right)^{x+ \frac{1}{2}} \geq \log(1)=e$since $\log$ is strictly increasing. ![[Pasted image 20240224231114.png]] *Question 2* Let $f(x)=\frac{1}{\sqrt{ x }} -\frac{1}{x}$. Then $f'(x)= x^{-2} - \frac{1}{2} x^{-\frac{3}{2}}$ If for $x\in(0,1)$, $f'(x)=0$ then $\frac{1}{x^{2}} - \frac{1}{2x^{\frac{3}{2}}} = 0 \implies \frac{2x^{\frac{3}{2}} -x^{2}}{2x^{\frac{7}{2}}} =0 \implies \frac{x^{\frac{3}{2}}\left( 2-x^{\frac{1}{2}} \right)}{2x^{\frac{7}{2}}} =0 \implies x=4 $since $x\neq 0$. Now $f''(x)= \frac{3}{4} x^{-\frac{5}{2}} -2x^{-3}$ so $f''(4)= \frac{3}{4}\cdot \frac{1}{2^{5}} -\frac{2}{4^{3}}= \frac{3}{4} \cdot \frac{1}{2^{5}}-\frac{1}{2^{5}}\leq 0$. So $f(4)$ is the maximum value of $f$ on $(0, \infty)$. Also $f(4)=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$. *Question 3* The derivative of $x \mapsto \log x$ is $x \mapsto \frac{1}{x}$ so $\lim_{ h \to 0 } \frac{\log(1+h)-\log(1)}{h} = 1$For each $t$, let $h=\frac{t}{n}$ then as $h\to{0}$, $n\to \infty$ and $\log(1)=0$ so $\lim_{ n \to \infty } \frac{n}{t} \log\left( 1+\frac{t}{n} \right) =1$