Chapter 2.6, Problem 36E

To determine

the value of the function $f$ approaches as the magnitude of $x$ increases

Answer to Problem 36E

The $f\left(x\right)$ approaches to $0$ , the value of $f\left(x\right)$ is greater and the solution is less

Explanation of Solution

Given information : The function is $f\left(x\right)=\frac{2x-1}{x^2+1}$

Calculation :

For the solution of $f\left(x\right)$

  $\begin{array}{l}f\left(x\right)=\frac{2x-1}{x^2+1}\\ \text{as }x\to \infty ,\frac{2x-1}{x^2+1}\to 0\end{array}$

At very large value of $x$ , the function approaches $\frac{2\infty -1}{{\infty }^2+1}$ which is equal to $0$

  $\begin{array}{l}\frac{199}{10,001}=0.02\\ \frac{-201}{10,001}=-0.02\end{array}$

At very large value of $x$ , $f\left(x\right)$ would be positive and approaches to. Thus, $f\left(x\right)$ would be more than the asymptote. When looking at values of $x$ approaching $-\infty$ , the numerator is negative while the denominator is positive, giving us a negative value