Q. 71 - 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 $a_k\to L$and $a_k\to M$, then $L=M$.

Step-by-Step Solution

Answer

The theorem has been proved.

Step 1. Given Information

The objective is to prove that $L=M$.

Step 2. Forming the equation.

The sequence $\left\{a_k\right\}$ is convergent such that $a_k\to L$.

By definition, for given $\epsilon >0$,there exists a positive integer $N$ such that

$\left|a_k-L\right|<\epsilon$ for $k⩾N.........\left(1\right)$

The sequence $\left\{a_k\right\}$ is convergent such that $a_k\to M$.

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

$\left|a_k-M\right|<\epsilon$ for $k\ge P.......\left(2\right)$

Step 3. Proving the theorem

Choose $R=max\left\{N,P\right\}$

Therefore,

$$\begin{gathered} \left|L-M\right|=\left|a_k-a_k+L-M\right| \\ =\left|L-a_k-(M-a_k)\right| \\ \le \left|a_k-L\right|+\left|a_k-M\right| \\ <\epsilon +\epsilon \\ =2\epsilon \end{gathered}$$

The inequality $\left|L-M\right|<2\epsilon$ holds for every $\epsilon >0$, and $\left|L-M\right|$ is independent of $\epsilon$.

Therefore,

$$\begin{gathered} \left|L-M\right|=0 \\ L=M \end{gathered}$$

Hence, proved.