5.1 Sequences  (Page 5/25)

Proof

We prove part iii.

Let $ϵ>0.$ Since $\underset{n\to \infty }{\text{lim}}{a}_n=A,$ there exists a constant positive integer $N_1$ such that for all $n\ge N_1.$ Since $\underset{n\to \infty }{\text{lim}}{b}_n=B,$ there exists a constant $N_2$ such that $|b_n-B|<\epsilon \text{/}2$ for all $n\ge N_2.$ Let $N$ be the largest of $N_1$ and $N_2.$ Therefore, for all $n\ge N,$

$|(a_n+b_n)\text{−}(A+B)|\le |a_n-A|+|b_n-B|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$

□

The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence $\left\{\frac{1}{n^2}\right\}.$ As shown earlier, $\underset{n\to \infty }{\text{lim}}1\text{/}n=0.$ Similarly, for any positive integer $k,$ we can conclude that

In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.

Recall that if $f$ is a continuous function at a value $L,$ then $f\left(x\right)\to f(L)$ as $x\to L.$ This idea applies to sequences as well. Suppose a sequence $a_n\to L,$ and a function $f$ is continuous at $L.$ Then $f\left(a_n\right)\to f\left(L\right).$ This property often enables us to find limits for complicated sequences. For example, consider the sequence $\sqrt{5-\frac{3}{n^2}}.$ From [link] a. we know the sequence $5-\frac{3}{n^2}\to 5.$ Since $\sqrt{x}$ is a continuous function at $x=5,$