Q. 88 Prove the second part of Theorem... [FREE SOLUTION]

Chapter 1: Q. 88 (page 151)

Prove the second part of Theorem $1.30$ : If $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)}$ is of the form $\frac{1}{- 鈭�}$ , then $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)} = 0$ .

Short Answer

Expert verified

It is proved thatIf $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)}$ is of the form $\frac{1}{- 鈭�}$ , then $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)} = 0$ .

Step by step solution

Step 1. Given Information

We are given two functions $f (x) and g (x)$ .

Step 2. Proving the statement

Consider a function $f (x)$ approaches $1$ that is $f (x) 鈫� 1$ .

For any $蔚 > 0$ , the function $f (x)$ satisfies that $1 - 蔚 < f (x) < 1 + 蔚$ .

Consider a function $g (x)$ approaches $- 鈭�$ that is $g (x) 鈫� - 鈭�$ .

Consider a number M which is less than $0$ that is $M < 0$ . Therefore, the function $g (x)$ can be written as $g (x) < M$ .

The number can be written as $M = \frac{- 2}{蔚}$ and consider $蔚 = 1$ .

Divide the function $f (x)$ by the function $g (x)$ ,

$\frac{1 + 蔚}{M} < \frac{f (x)}{g (x)} < 1 - 蔚 \frac{1 + 蔚}{(- \frac{2}{蔚})} < \frac{f (x)}{g (x)} < 1 - 蔚 \frac{1 + 1}{(- \frac{2}{1})} < \frac{f (x)}{g (x)} < 1 - 1 - 1 < \frac{f (x)}{g (x)} < 0$

Hence, if $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)}$ is in the form of $\frac{1}{- 鈭�}$ then $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)} = 0$ . Hence the given statement is proved.

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91影视

Prove the second part of Theorem $1.30$ : If $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)}$ is of the form $\frac{1}{- 鈭�}$ , then $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)} = 0$ .

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

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91影视

Prove the second part of Theorem 1.30: If limx鈫�鈭�f(x)g(x)is of the form 1-鈭�, then limx鈫�鈭�f(x)g(x)=0.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Probability and Statistics

Applied Mathematics

Decision Maths

Theoretical and Mathematical Physics

Geometry

Study anywhere. Anytime. Across all devices.

Prove the second part of Theorem $1.30$ : If $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)}$ is of the form $\frac{1}{- 鈭�}$ , then $\lim_{x 鈫� 鈭�} \frac{f (x)}{g (x)} = 0$ .