Chapter 1: Q. 88 (page 151)

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

Short Answer

Expert verified

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

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 $f (x) \to 1$ .

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

Consider a function $g (x)$ approaches $- \infty$ that is $g (x) \to - \infty$ .

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 $M = \frac{- 2}{ε}$ and consider $ε = 1$ .

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

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

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

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

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

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Prove the second part of Theorem 1.30: If limx→∞f(x)g(x)is of the form 1-∞, then limx→∞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

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Probability and Statistics

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