Q. 69 - Calculus

Question

Let $\left\{a_k\right\}$ and $\left\{b_k\right\}$ be convergent sequences with $a_k\to L$ and $b_k\to M$ as $k\to \infty$ and let $c$ be a constant. Prove the indicated basic limit rules from Theorem 7.11. You may wish to model your proofs on the proofs of the analogous statements from Section 1.5.

Prove that $a_k{b}_k\to LM$.

Step-by-Step Solution

Answer

Hence, the theorem is proved.

Step 1. Given Information.

The objective is to prove that $a_k{b}_k\to LM$.

Step 2. Forming the equations.

We use the definition of convergence for the sequence $\left\{a_k\right\}$ and $\left\{b_k\right\}$.

The sequence $\left\{a_k\right\}$ converges to $L$.

For given $\epsilon >0$, there exists a positive integer  such that

$\left|a_k-L\right|<\frac{\epsilon }{2m}$ for $k\ge N..........\left(1\right)$

The sequence $\left\{b_k\right\}$ converges to $M$.

For given $\epsilon >0$, there exists a positive integer $P$ such that

$\left|b_k-M\right|<\frac{\epsilon }{2\left|L\right|+1}$ for $k\ge P...........\left(2\right)$

There exists a positive integer $m$ such that

$\left|b_k\right|\le m$ for all $k...........\left(3\right)$

Step 3. Proving the theorem.

Choose a positive integer $R$ such that $R=max\left\{N,P\right\}$

$$\begin{gathered} \left|a_k{b}_k-LM\right|=\left|\left(a_k-L\right)b_k+\left(b_k-M\right)L\right| \\ \le \left|\left(a_k-L\right)\right|\left|b_k\right|+\left|\left(b_k-M\right)\right|\left|L\right| \\ <\frac{\epsilon }{2m}\times m+\frac{\epsilon }{2\left(\left|L\right|+1\right)}\times L for k\ge R \\ <\frac{\epsilon }{2}+\frac{\epsilon }{2} \\ =\epsilon \end{gathered}$$

Thus, for $k\ge R,\left|a_k{b}_k-\left(LM\right)\right|<\epsilon$ and hence, $\left\{a_k{b}_k\right\}$ is convergent.

Therefore, the value is $\underset{k\to \infty }{\lim }\left\{a_k{b}_k\right\}=LM$.