Answer
By graphing the function on the interval $-10 \leq x \leq 10$, we would guess that the horizontal asymptote is $y = 1$
$\lim\limits_{x \to \infty}f(x) = 3$
$\lim\limits_{x \to -\infty}f(x) = 3$
The horizontal asymptote is $y = 3$
The interval $-10 \leq x \leq 10$ does not give us the true indication of the horizontal asymptote because this function approaches the asymptote $y=3$ very gradually.
Work Step by Step
$f(x) = \frac{3x^3+500x^2}{x^3+500x^2+100x+2000}$
By graphing the function on the interval $-10 \leq x \leq 10$, we would guess that the horizontal asymptote is $y = 1$
We can evaluate the limit as $x$ approaches infinity:
$\lim\limits_{x \to \infty}\frac{3x^3+500x^2}{x^3+500x^2+100x+2000}$
$= \lim\limits_{x \to \infty}\frac{3x^3/x^3+500x^2/x^3}{x^3/x^3+500x^2/x^3+100x/x^3+2000/x^3}$
$= \frac{3}{1}$
$= 3$
Similarly, $\lim\limits_{x \to -\infty}f(x) = 3$
The horizontal asymptote is $y = 3$
The interval $-10 \leq x \leq 10$ does not give us the true indication of the horizontal asymptote because this function approaches the asymptote $y=3$ very gradually.
