Question

2. Sequence Convergence¶ (To be answered in LaTex) Show that the following sequence limits using either...

2. Sequence Convergence¶ (To be answered in LaTex)

Show that the following sequence limits using either an ?−?ε−δ argument

a) lim (3?+1) / (2?+5) = 3/2 answer in LaTex

b) lim (2?) / (n+2) =2   answer in LaTex

c) lim [(1/?) − 1/(?+1)] = 0 answer in LaTex

(a) Let $\epsilon >0$, then by Archimedian property there exists $N \in \mathbf{N}$ such that for all $n > N$ we have $\frac{1}{2n+5} < \frac{2 \epsilon}{5}$.

Now $|\frac{3n+1}{2n+5}- \frac{3}{2}| = |\frac{13}{2(2n+5)}| < \epsilon$ for all $n > N$. Hence $\lim \frac{3n+1}{2n+5} = \frac{3}{2}$.

(b) Let $\epsilon >0$, then by Archimedian property there exists $N \in \mathbf{N}$ such that for all $n > N$ we have $\frac{1}{n+2} < \frac{ \epsilon}{4}$.

Now $|\frac{2n}{n+2}- 2| = |\frac{4}{n+2}| < \epsilon$ for all $n > N$. Hence $\lim \frac{2n}{n+2} = 2$.

(c) Let $\epsilon >0$, then by Archimedian property there exists $N \in \mathbf{N}$ such that for all $n > N$ we have $\frac{1}{n} < \frac{ \epsilon}{2}$.

Now $|\frac{1}{n}- \frac{1}{n+1}| < |\frac{1}{n}|+ |\frac{1}{n+1}| < \epsilon$ for all $n > N$. Hence $\lim \frac{1}{n}- \frac{1}{n+1} = 0$.