Q. 70 - 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 if $M\ne 0$, then $\frac{a_k}{b_k}\to \frac{L}{M}$.

Step-by-Step Solution

Answer

Hence, the theorem is proved.

Step 1. Given Information.

The objective is to prove that $\frac{a_k}{b_k}\to \frac{L}{M}$

Step 2.Proving the theorem.

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

The value of $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$ is,

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\left(a_k\times \frac{1}{b_k}\right) \\ =\left(\underset{k\to \infty }{\lim }{a}_k\right)\left(\underset{k\to \infty }{\lim }\frac{1}{b_k}\right) \\ =L\left(\underset{k\to \infty }{\lim }\frac{1}{b_k}\right)..............\left(1\right) \end{gathered}$$

The reciprocal rule states that if $\underset{x\to \infty }{\lim }{c}_x=c$ with $c\ne 0$, then,

$\underset{x\to \infty }{\lim }\frac{1}{c_x}=\frac{1}{c}$

Therefore,

$\underset{x\to \infty }{\lim }\frac{1}{b_k}=\frac{1}{M}$ (because $b_k\to M$)

Therefore, equation(1) is written as,

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=L\left(\frac{1}{M}\right) \\ =\frac{L}{M} \end{gathered}$$

Therefore, hence proved.