AP Calculus AB and BC

Limit of x^2/(x+1) as x Approaches Infinity

Q: What is the limit as x approaches negative infinity?

It is -\infty. The same degree comparison applies, and the surviving factor of x is negative there, so the values run down without bound. Concretely, the function tracks y = x - 1 on both ends.

Q: Does this function have a vertical asymptote too?

Yes, at x = -1, where the denominator is 0 and the numerator is 1. That is a separate question from end behavior: vertical asymptotes come from finite trouble spots, horizontal and slant asymptotes come from limits at infinity.

The limit of x squared over x plus 1 as x approaches infinity is infinity. Degree 2 on top beats degree 1 on the bottom, leaving one factor of x, so the values grow without bound. There is no horizontal asymptote, since the values never settle on a finite number; the end behavior follows the slant asymptote y = x - 1.

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

Settled by comparing leading degrees.

Reading the degrees off the ratio

End behavior of a rational function is decided by its leading terms, because every lower-order piece becomes negligible once $x$ is large. The numerator here leads with $x^2$ and the denominator leads with $x$ , so the ratio behaves like $x$ itself.

Dividing every term by $x$ , the highest power in the denominator, turns that observation into algebra you can defend.

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

As $x \to \infty$ the term $\frac{1}{x}$ dies to 0, so the denominator settles at 1 while the numerator runs off unbounded. A quantity growing like $x$ over a quantity approaching 1 is unbounded.

\lim_{x \to \infty} \frac{x}{1 + \frac{1}{x}} = \infty

Polynomial division confirms the same shape and shows exactly how fast the growth is. Since $x^2 = (x+1)(x-1) + 1$ , the function splits into a linear part and a vanishing remainder.

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

The remainder $\frac{1}{x+1}$ goes to 0, so for large $x$ the function tracks the line $y = x - 1$ and inherits its unbounded growth.

Why substitution gives you nothing

The reflex is to plug in, but $\infty$ is not a number you can substitute. Pushing it through anyway produces $\frac{\infty}{\infty}$ , which is indeterminate: it records that both parts are unbounded and says nothing about which one wins.

That the form decides nothing is easy to see. Three quotients share it and reach three different answers.

Quotient	Form at infinity	Limit
$\frac{x}{x^2}$	$\frac{\infty}{\infty}$	$0$
$\frac{3x}{x}$	$\frac{\infty}{\infty}$	$3$
$\frac{x^2}{x}$	$\frac{\infty}{\infty}$	$\infty$

L'Hopital's rule is legal on this form and does confirm the answer, since differentiating top and bottom gives $\frac{2x}{1}$ , which is unbounded. It is slower than the degree comparison and offers no extra information, which is why the degree count is the technique to reach for on a rational function.

The mistake students make

The common wrong answer is 1, from cancelling an $x$ and reading off the leading coefficients. That shortcut belongs to a different case: the ratio of leading coefficients is the limit only when the two degrees are equal, and here they are not.

Degree comparison	Limit at infinity
Numerator degree less than denominator degree	$0$ , with horizontal asymptote $y = 0$
Degrees equal	Ratio of the leading coefficients
Numerator degree greater	Unbounded, so no horizontal asymptote

No horizontal asymptote does not mean no asymptote

A horizontal asymptote needs a finite limit at infinity, so this function has none. It does have the slant asymptote $y = x - 1$ , which comes straight out of the division above. When the numerator degree beats the denominator degree by exactly one, expect a slant asymptote.

The other slip is answering $0$ because $\frac{1}{x+1}$ appears in the divided form and looks small. That term is the remainder, not the function.

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

If the limit is infinity, does the limit exist?

Strictly, no: no finite number is being approached, so the limit fails to exist. Writing $\infty$ is the standard way of naming how it fails, which is that the values increase without bound. On AP work write $\infty$ rather than a bare 'does not exist', since the symbol carries the extra information about direction.

What is the limit as x approaches negative infinity?

It is $-\infty$ . The same degree comparison applies, and the surviving factor of $x$ is negative there, so the values run down without bound. Concretely, the function tracks $y = x - 1$ on both ends.

Does this function have a vertical asymptote too?

Yes, at $x = -1$ , where the denominator is 0 and the numerator is 1. That is a separate question from end behavior: vertical asymptotes come from finite trouble spots, horizontal and slant asymptotes come from limits at infinity.