Verify that $|\frac{2^n-1}{2^n+1}-1|<mn>0</mn><mn>.</mn><mn>0</mn><mn>0</mn><mn>0</mn><mn>1</mn>$ for all sufficiently large values $ofn.$ How large $is$

"sufficiently large"? Hint: Show that $|\frac{2^n-1}{2^n+1}-1|=\frac{2}{2^n+1}.$

Limiting value $of$ a sequence. $Ifpis$ any small positive number, find $N$ such that for all

$n>N$ the following inequality $is$ true: $\underset{n}{lim}\frac{2^n-1}{2^n+1}=1|\frac{2^n-1}{2^n+1}-1|.Itis$ customary $to$ summarize this

situation $by$ writing $\underset{n}{lim}\frac{2^n-1}{2^n+1}=1.$