a$-$ Let $A=\{1-\frac{1}{n},$ninN$\}.$

Prove that supA$=1.$

b$-$ Let a and $b$ be real numbers with $f$:$[a,b](0,)f(a,b)cin(a,b)\frac{f(a)}{f(b)}=$exp$((a-b)\frac{f^{\text{'}}(c)}{f(c)}).\underset{x1}{lim}\frac{2x^2-1}{x-2}=-1$;$\underset{x}{lim}(\sqrt[\phantom{2}]{x+1}-\sqrt[\phantom{2}]{x-1})=0$ARa$=$supA$\{a_n{\}}_{\text{n}\text{i}\text{n}\text{N}}A\underset{n}{lim}{a}_n=$aARdelA$=\frac{O}{?}A(x,||*|{|}_x)(Y,||*|{|}_Y)||*||$:$xYR||*||xY{a}_n=\sqrt[\phantom{2}]{n}-1,$ninN${\displaystyle \underset{n=1}{\overset{}{}}}n(\sqrt[\phantom{2}]{1+\frac{n+1}{n^2}}-\sqrt[\phantom{2}]{1+\frac{-n+1}{n^2}})$

${\displaystyle \underset{n=1}{\overset{}{}}}(1+\frac{1}{n}{)}^{n^2}{\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{n(n+1{)}^2}{\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{n^2}=\frac{{}^2}{6}KRf$:$KRKf(K)(lon,)\underset{x0}{lim}x+\sqrt[\phantom{2}]{x^2+1}=1f-f(x)=x^3-x^2+1f(-1)f(1)f(x)=-\frac{1}{2}(-1,1)$