Question

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

Answer 1

(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 $\lim \frac{1}{n}- \frac{1}{n+1} = 0$.