Question

A sequence is a function whose domain is the set of natural number. A sequence $s_1,s_2...$ of real numbers is said to have a limit $l$ if ${lim}_{n\to \infty }{s}_n=l$. If $l<\infty ,$ then $<s_n>$ is said to convergent. The following are well known
(i) $\underset{n\to \infty }{lim}\frac{1}{n^p}=0(p>0)$

(ii) $\underset{n\to \infty }{lim}{x}^n=0(\left|x\right|<1)$

(iii) If $\underset{n\to \infty }{lim}{s}_n=l$, then $\underset{n\to \infty }{lim}\frac{s_1+s_2+...+s_n}{n}=l$

Let $f\left(x\right)=-1+|x-1|,g\left(x\right)=2-|x+1|$ then

• A. $\underset{n\to \infty }{lim}fog\left(1/n\right)=0$
• B. $\underset{n\to \infty }{lim}fog\left(1/n\right)=-1$
• C. $\underset{n\to \infty }{lim}gof\left(1/n\right)=0$
• D. $\underset{n\to \infty }{lim}gof\left(1/n\right)=-1$

Answer 1

The correct option is B $\underset{n\to \infty }{lim}fog\left(1/n\right)=-1$

$fog\left(x\right)=\{\begin{array}{cc}-(x+3),& x<-2\\ x+1,& -2\le x\le -1\\ -1-x,& -1\le x\le 0\\ x-1,& x>0\end{array}$

$gof\left(x\right)=\{\begin{array}{cc}1+x,& x<-1\\ 1-x,& -1\le x\le 0\\ 3-x,& x>0\end{array}$

$fog\left(\frac{1}{n}\right)=\frac{1}{n}-1\to =-1$ as $n\to \infty$

and

$gof\left(\frac{1}{n}\right)=1-\frac{1}{n}\to =1$ as $n\to \infty$