Chapter D, Problem 20E

(a)

To determine

To calculate:

Values of $x$ .

Answer to Problem 20E

Value of $x<0.001$ .

Explanation of Solution

Given information:

  $\frac{1}{x^2}>1000000$

Calculation:

Let $f$ be a function defined in some interval $\left(a,\infty \right)$

  $\underset{x\to \infty }{\lim }f\left(x\right)=L$

Means that for every number $\epsilon >0$ there is corresponding number $N$ such that

If $x>N$ then $\left|f\left(x\right)-L\right|<\epsilon$

The given inequality is:

  $\frac{1}{x^2}>1000000$

So the value of $x$ is:

  $\begin{array}{l}{x}^2<\frac{1}{1000000}\\ x^2<0.000001\\ x<0.001\\ x<0.001\end{array}$

(b)

To determine

To prove:

  ${\lim }_{x\to 0}\left(1/x^2\right)=\infty$ by using definition.

Answer to Problem 20E

Hence, $\underset{x\to 0}{\lim }\frac{1}{x^2}=\infty$ is proved.

Explanation of Solution

Given information:

  $\frac{1}{x^2}>1000000$

  $0<\left|x-a\right|<\delta$

Proof:

Here,

  $f\left(x\right)=\frac{1}{x^2},a=0,L=\infty$

Now from the precise definition of limit for every positive number $M$ there is a corresponding positive number $\delta$ such that if $0<\left|x-a\right|<\delta$ then $f\left(x\right)>M$ .

Thus, if $0<\left|x-a\right|<\delta$ then $\frac{1}{x^2}>M$ or $\left|x\right|<\frac{1}{\sqrt{M}}$ .

One can select $\delta =\frac{1}{\sqrt{M}}$ then $x^2<\frac{1}{M}\iff \left|x\right|<\delta$ .

Therefore, one can say that

  $\underset{x\to 0}{\lim }\frac{1}{x^2}=\infty$