Chapter 2.5, Problem 45E

To determine

To estimate:the horizontal asymptotes of the given function by graphing f for $10\le x\le 10$ .

Answer to Problem 45E

  $f\left(x\right)=\frac{3x^3+500x^2}{x^3+500x^2+100x+2000}=3$

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{3x^3+500x^2}{x^3+500x^2+100x+2000}$ .

Calculation:

An estimate of the horizontal asymptote is the line $y=1$ .

Evaluating the limit, it is found that

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{3x^3+500x^2}{x^3+500x^2+100x+2000}\\ =\underset{x\to \infty }{\lim }\frac{x^3\left(3+\frac{500}{x}\right)}{x^3\left(3+\frac{500}{x}+\frac{100}{x^2}+\frac{2000}{x^3}\right)}\\ =\underset{x\to \infty }{\lim }\frac{\left(3+\frac{500}{x}\right)}{\left(3+\frac{500}{x}+\frac{100}{x^2}+\frac{2000}{x^3}\right)}\\ =\frac{\left(3+0\right)}{\left(1+0+0+0\right)}=3\end{array}$

The reason for this dispensary is that the limits of the graph are too small to show the behavior of the function for large values of x .

Now, graph the above function: