Q. 73 - Calculus

Question

Let $\left\{a_k\right\}$ be a sequence. Prove the indicated limit rules from Theorem 7.12. You may wish to model your proofs on the proofs of the analogous statements from Section 1.5.

Prove that if $\left\{m_k\right\}$ and $\left\{M_k\right\}$ are sequences that both converge to $L$, and if $m_k\le a_k\le M_k$ for all $k$, then $a_k\to L$.

Step-by-Step Solution

Answer

The theorem is proved.

Step 1. Given Information.

The objective is to prove that $a_k\to L$.

Step 2. Forming the equation.

The sequence $\left\{m_k\right\}$ is convergent and converges to $L$.

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

$$\begin{gathered} \left|m_k-L\right|<\epsilon , k\ge N \\ L-\epsilon <m_k<\epsilon +L for k\ge N..........\left(1\right) \end{gathered}$$

The sequence $\left\{M_k\right\}$ is convergent and converges to $L$.

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

$$\begin{gathered} \left|M_k-L\right|<\epsilon ,k\ge M \\ L-\epsilon <M_k<\epsilon +L for k\ge M...........\left(2\right) \end{gathered}$$

Step 3. Proving the theorem.

It is given that $m_{k\le }{a}_k\le M_k$

Choose $P=max(N,M)$

Using equation(1) and (2) we get,

$$\begin{gathered} L-\epsilon <m_k\le a_k\le M_k<L+\epsilon for k\ge P \\ L-\epsilon <a_k<L+\epsilon for k\ge P \\ \left|a_k-L\right|<\epsilon for k\ge P \end{gathered}$$

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

$\left|a_k-L\right|<\epsilon for k\ge P$.

Therefore, the sequence $\left\{a_k\right\}$converges to limit $L$.