AP Calculus AB and BC

Limit of (5x^3-2x)/(3x^3+7) at Infinity Is 5/3

Q: Does L'Hopital's rule reach 5/3 as well?

Yes, in three passes: \frac{15x^2-2}{9x^2}, then \frac{30x}{18x}, then \frac{30}{18} = \frac{5}{3}. You can also cancel at the middle stage. For rational functions the division is faster, and it hands you the horizontal asymptote in the same step.

Q: Is the limit still 5/3 as x approaches negative infinity?

Yes. Writing the fraction as \frac{5 - \frac{2}{x^2}}{3 + \frac{7}{x^3}} works for negative x too, and both small terms still go to 0. So y = \frac{5}{3} is the horizontal asymptote at both ends of the graph.

Q: Why do the -2x and the +7 not matter?

They are swamped. At x = 1000 the term 5x^3 is five billion while 2x is two thousand, so the numerator is 5x^3 to nine significant figures. Low-order terms decide the shape near the origin, not the end behaviour.

The limit of (5x^3 - 2x)/(3x^3 + 7) as x approaches infinity is 5/3, about 1.667. Substitution gives infinity over infinity. Divide top and bottom by x cubed and the leftover terms vanish, leaving the ratio of the leading coefficients, 5 over 3, which is also the horizontal asymptote.

\lim_{x \to \infty} \frac{5x^3-2x}{3x^3+7} = \frac{5}{3}

Settled by comparing leading degrees.

Dividing by the dominant power

The highest power anywhere in the fraction is $x^3$ , so divide every term above and below by it. Dividing the top and the bottom by the same nonzero quantity leaves the value untouched.

\frac{5x^3-2x}{3x^3+7} = \frac{5 - \frac{2}{x^2}}{3 + \frac{7}{x^3}}

As $x \to \infty$ both small fractions collapse to $0$ and only the leading coefficients are left standing.

\lim_{x \to \infty} \frac{5 - \frac{2}{x^2}}{3 + \frac{7}{x^3}} = \frac{5-0}{3+0} = \frac{5}{3}

The approach is from below, because the $-\frac{2}{x^2}$ trims the top while the $+\frac{7}{x^3}$ pads the bottom.

$x$	$\frac{5x^3-2x}{3x^3+7}$
$10$	$1.65614$
$100$	$1.66660$
$1000$	$1.6666660$

For comparison, $\frac{5}{3} = 1.66667$ to five places.

Why substitution fails

Both cubics run to $\infty$ , so substitution returns $\frac{\infty}{\infty}$ , and that form is silent on which one runs faster. Change one exponent and the answer moves anywhere: $\frac{5x^3-2x}{3x^4+7} \to 0$ and $\frac{5x^4-2x}{3x^3+7} \to \infty$ , while the form stays identical.

Growth rate is what settles it, and for a polynomial the growth rate is its degree. Dividing by the dominant power is the standard way of putting that comparison on the page instead of in your head.

Reading the answer off the degrees

Numerator degree lower than denominator degree: the limit is $0$ , and $y = 0$ is the horizontal asymptote.
Degrees equal: the limit is the ratio of the leading coefficients, and that ratio is the horizontal asymptote.
Numerator degree higher: the limit is $\infty$ or $-\infty$ , and there is no horizontal asymptote.

Both polynomials here have degree $3$ , so the middle case applies and $y = \frac{5}{3}$ is the horizontal asymptote. The $-2x$ and the $+7$ shape the graph near the origin and get no vote at the far end.

The same value holds in the other direction. Odd powers do flip sign, but they flip on the top and the bottom together, so the two sign changes cancel inside the ratio.

\lim_{x \to -\infty} \frac{5x^3-2x}{3x^3+7} = \frac{5}{3}

One vertical asymptote, irrelevant here

The denominator does vanish once, at $x = -\sqrt[3]{\frac{7}{3}} \approx -1.326$ , where the numerator is about $-9.01$ rather than $0$ . That gives a genuine vertical asymptote, and it says nothing at all about end behaviour. Vertical asymptotes answer questions at finite inputs; the horizontal one answers this question.

The mistakes students make

Crossing out $5x^3$ against $3x^3$ as though the terms cancel. Terms of a sum never cancel across a fraction bar; dividing by $x^3$ is the move that makes the comparison legal.
Combining all four coefficients, as in $\frac{5-2}{3+7}$ . Only the leading pair survives the division, and the rest go to $0$ .
Answering $\frac{3}{5}$ . The numerator's coefficient stays on top, exactly where it started.
Assuming an odd degree forces a different answer at $-\infty$ . That happens for a single polynomial, not for a ratio of two odd-degree polynomials, where the signs cancel.
Treating $x \approx -1.326$ as a problem. The limit at infinity never looks at any finite input.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Does L'Hopital's rule reach 5/3 as well?

Yes, in three passes: $\frac{15x^2-2}{9x^2}$ , then $\frac{30x}{18x}$ , then $\frac{30}{18} = \frac{5}{3}$ . You can also cancel at the middle stage. For rational functions the division is faster, and it hands you the horizontal asymptote in the same step.

Is the limit still 5/3 as x approaches negative infinity?

Yes. Writing the fraction as $\frac{5 - \frac{2}{x^2}}{3 + \frac{7}{x^3}}$ works for negative $x$ too, and both small terms still go to $0$ . So $y = \frac{5}{3}$ is the horizontal asymptote at both ends of the graph.

Why do the -2x and the +7 not matter?

They are swamped. At $x = 1000$ the term $5x^3$ is five billion while $2x$ is two thousand, so the numerator is $5x^3$ to nine significant figures. Low-order terms decide the shape near the origin, not the end behaviour.