Chapter D, Problem 17E

To determine

To calculate:

Value of $N$ by using graph.

Answer to Problem 17E

For $\epsilon =0.2$ value of $N$ .is $13$ .

Explanation of Solution

Given information:

If $x>N$ then $\left|\frac{6x^2+5x-3}{2x^2-1}-3\right|<0.2$

  $\epsilon =0.5,\epsilon =0.1$

Calculation:

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

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

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

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

Given function is:

  $f\left(x\right)=\frac{6x^2+5x-3}{2x^2-1}$

The graph $f\left(x\right)=\frac{6x^2+5x-3}{2x^2-1}$ and its corresponding values are shown as follows:

  

For finding out $N$ ,

If $x>N$ then $\left|\frac{6x^2+5x-3}{2x^2-1}-3\right|<0.2$

From the inequality,

  $\begin{array}{l}-0.2<\frac{6x^2+5x-3}{2x^2-1}-3<0.2\\ 2.8<\frac{6x^2+5x-3}{2x^2-1}<3.2\end{array}$

Here required to find out value of $x$ for which given curve lies between the horizontal lines $y=2.8,y=3.2$ .

  

Here cursor is used to estimate that the curve crosses the line $y=3.2$ when $x=12.74$ as shown.

To right of number curve stays between the line $y=2.8,y=3.2$ and thus cannot round of the value of $x$ .

If $x>13$ then $\left|\frac{6x^2+5x-3}{2x^2-1}-3\right|<0.2$

Hence, For $\epsilon =0.2$

  $N=13$ .