Suppose that a friend tells you that if any number M > 0 were to be speci-fied, one would always be able to find a corresponding number 8 > 0 such thatthe relation > M holds for all x € (-d,0). State if you agree or disagree withthe friend. Motivate briefly.

Answered: Suppose that a friend tells you that if…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Suppose that a friend tells you that if any number M > 0 were to be speci-
fied, one would always be able to find a corresponding number 8 > 0 such that
the relation > M holds for all x € (-d,0). State if you agree or disagree with
the friend. Motivate briefly.

Expert Solution

Step 1

The friend tells that if any number $M > 0$ is specified, it always can be able to find a corresponding number $δ > 0$ such that the relation $\frac{1}{x} > M$ holds for all $x \in (- δ, 0)$ .

The statement is wrong.

It is given that $M > 0$ .

Let $δ > 0$ . Multiply the inequality by $- 1$ and change the sign of the inequality:

$\begin{array}{rcl} (- 1) δ & < & (- 1) 0 \\ - δ & < & 0 \end{array}$

If $x \in (- δ, 0)$ , then it follows:

$- δ < x < 0$

Note that $x < 0$ . So the value of $x$ never be $0$ . The expression $x^{2}$ is always positive if $x \neq 0$ . If an inequality is multiplied by a positive constant, the sign will not change. So multiply the above inequality by $x^{2}$ :

$\begin{array}{rcl} \frac{- δ}{x^{2}} & < & \frac{x}{x^{2}} < \frac{0}{x^{2}} \\ \frac{- δ}{x^{2}} & < & \frac{1}{x} < 0 \end{array}$

Note that $M > 0$ . It follows:

$\frac{1}{x} < 0 < M \frac{1}{x} < M \frac{1}{x} ≯ M$

Hence, for every $δ > 0$ , if $x \in (- δ, 0)$ , then $\frac{1}{x} ≯ M$ .

For a given $M > 0$ , it is not possible to choose $δ > 0$ such that $x \in (- δ, 0)$ , then $\frac{1}{x} > M$ . So, the statement is wrong, the statement has to be disagreed.