AP Calculus AB and BC

Limit of (2x^2+3)/(x^2-1) at Infinity

Q: Does the limit change as x approaches negative infinity?

No, it is still 2. The dominant terms 2x^2 and x^2 are even powers, so their ratio does not care about the sign of x, and y = 2 is the horizontal asymptote in both directions.

Q: Can I just use L'Hopital?

Yes, twice, ending at \frac{4}{2} = 2. Dividing by x^2 is usually quicker for rational functions and it exposes the horizontal asymptote at the same time.

Q: Why do the constants 3 and -1 not matter?

At large x they are swamped. At x = 1000 the term 2x^2 is two million while the +3 is a rounding error on it. Constants shape the graph near the origin, not the end behaviour.

The limit of (2x^2 + 3)/(x^2 - 1) as x approaches infinity is 2. Substitution gives infinity over infinity, so divide the top and bottom by x squared and the leftover terms vanish. Equal degrees means the answer is the ratio of leading coefficients, 2 over 1, and y = 2 is the horizontal asymptote.

\lim_{x \to \infty} \frac{2x^2+3}{x^2-1} = 2

Settled by comparing leading degrees.

Dividing by the dominant power

Divide every term on the top and the bottom by $x^2$ , the highest power present. The value is untouched, since the same division is applied to both parts of the fraction.

\frac{2x^2+3}{x^2-1} = \frac{2 + \frac{3}{x^2}}{1 - \frac{1}{x^2}}

As $x \to \infty$ both $\frac{3}{x^2}$ and $\frac{1}{x^2}$ go to 0, leaving nothing but the constants.

\lim_{x \to \infty} \frac{2 + \frac{3}{x^2}}{1 - \frac{1}{x^2}} = \frac{2+0}{1-0} = 2

The numbers agree: at $x = 100$ the function is about $2.0005$ and at $x = 1000$ about $2.000005$ , closing on 2 from above.

L'Hopital gets there too

Differentiating top and bottom gives $\frac{4x}{2x}$ , which is still $\frac{\infty}{\infty}$ as written; cancel it, or run a second round to reach $\frac{4}{2} = 2$ . For polynomials the division is faster, and it hands you the horizontal asymptote in the same step.

Why substitution fails

Substitution gives $\frac{\infty}{\infty}$ . Both parts are unbounded, and the form carries no information about which is bigger or by how much. Three functions share it and disagree completely: $\frac{2x^2+3}{x^2-1} \to 2$ , $\frac{2x^2+3}{x^3-1} \to 0$ , and $\frac{2x^3+3}{x^2-1} \to \infty$ .

What settles the question is growth rate, which for polynomials is the degree. Dividing by the dominant power is the standard way to make that comparison visible on the page.

The degree rule this is an instance of

Degrees	Limit as $x \to \infty$	Horizontal asymptote
Numerator lower	0	$y = 0$
Equal	Ratio of leading coefficients	$y$ equals that ratio
Numerator higher	$\infty$ or $-\infty$	None

Both degrees are 2 here, so the limit is $\frac{2}{1} = 2$ . The $+3$ and the $-1$ shape the graph near $x = \pm 1$ and have no vote at the far end.

The same value holds in the other direction, because the dominant terms are even powers: $\lim_{x \to -\infty} \frac{2x^2+3}{x^2-1} = 2$ as well, so $y = 2$ is the horizontal asymptote at both ends.

The mistake students make

Crossing out the $x^2$ terms as though they cancel. That lands on 2 by luck here and fails elsewhere, since $2x^2 + 3$ has no $x^2$ factor to cancel.
Worrying about $x = \pm 1$ . Those roots of the denominator are vertical asymptotes and say nothing about behaviour at infinity.
Assuming a graph can never touch its horizontal asymptote. This one never does, since setting the function equal to 2 gives $3 = -2$ , but crossing is allowed in general and other rational functions do it.
Stopping at $\frac{4x}{2x}$ after one round of L'Hopital and calling it stuck. Cancel to $2$ , or differentiate once more for $\frac{4}{2}$ .

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 the limit change as x approaches negative infinity?

No, it is still 2. The dominant terms $2x^2$ and $x^2$ are even powers, so their ratio does not care about the sign of $x$ , and $y = 2$ is the horizontal asymptote in both directions.

Can I just use L'Hopital?

Yes, twice, ending at $\frac{4}{2} = 2$ . Dividing by $x^2$ is usually quicker for rational functions and it exposes the horizontal asymptote at the same time.

Why do the constants 3 and -1 not matter?

At large $x$ they are swamped. At $x = 1000$ the term $2x^2$ is two million while the $+3$ is a rounding error on it. Constants shape the graph near the origin, not the end behaviour.