Chapter 7: Q. 70 (page 605)

Let $\{a_{k}\}$ and $\{b_{k}\}$ 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 $M \neq 0$ , then $\frac{a_{k}}{b_{k}} \to \frac{L}{M}$ .

Short Answer

Expert verified

Hence, the theorem is proved.

Step by step solution

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 $\{a_{k}\}$ and $\{b_{k}\}$ .

The value of $\lim_{k \to \infty} \frac{a_{k}}{b_{k}}$ is,

$\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \lim_{k \to \infty} (a_{k} \times \frac{1}{b_{k}}) = (\lim_{k \to \infty} a_{k}) (\lim_{k \to \infty} \frac{1}{b_{k}}) = L (\lim_{k \to \infty} \frac{1}{b_{k}}) . . . . . . . . . . . . . . (1)$

The reciprocal rule states that if $\lim_{x \to \infty} c_{x} = c$ with $c \neq 0$ , then,

$\lim_{x \to \infty} \frac{1}{c_{x}} = \frac{1}{c}$

Therefore,

$\lim_{x \to \infty} \frac{1}{b_{k}} = \frac{1}{M}$ (because $b_{k} \to M$ )

Therefore, equation(1) is written as,

$\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L (\frac{1}{M}) = \frac{L}{M}$

Therefore, hence proved.

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2.Proving the theorem.

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Let akand bkbe convergent sequences with ak→Land bk→Mas k→∞and let cbe 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≠0, then akbk→LM.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2.Proving the theorem.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Calculus

Discrete Mathematics

Decision Maths

Applied Mathematics

Probability and Statistics

Study anywhere. Anytime. Across all devices.