Limits at Infinity

Take the function f(x) = \dfrac{3x^2 + 5}{x^2 + 1} and plug in some large values of x.

The outputs are closing in on 3. Not 3 exactly — there is always a small correction — but the correction shrinks as x grows. By x = 10{,}000, you are within a hundred-millionth of 3.

Now try x = -10, x = -100, x = -1000. The same thing happens: the outputs approach 3 from just above 3. The function tends to 3 from both directions.

This means that far to the right and far to the left, the graph of f is nearly flat, hovering just above the line y = 3. That line is called a horizontal asymptote — a value the function approaches but never quite reaches (or, in some cases, reaches and crosses) as x goes to \pm\infty.

The question "what does f(x) approach as x grows forever?" is a limit at infinity.

Compare this with the limits you have seen before. An ordinary limit like \lim_{x \to 2} f(x) asks what happens as x closes in on a specific number. A limit at infinity asks what happens when x escapes — when it runs off to the right (or left) forever. The two ideas are cousins. Both ask the same question: what value does f(x) crowd toward? The difference is in what x is doing while you ask.

What the notation means

The notation

means: as x increases without bound, f(x) gets arbitrarily close to L. You can make f(x) as close to L as you like, as long as you take x large enough.

Similarly, \lim_{x \to -\infty} f(x) = L means the same thing as x decreases without bound (goes to the far left on the number line).

There is an important nuance: x \to \infty does not mean x "reaches" infinity. Infinity is not a number — you cannot add it, subtract it, or divide by it. The notation means x grows beyond every finite bound. You will never "arrive at infinity"; you are always on your way. The limit L, if it exists, is the number the outputs crowd toward as the inputs keep growing — the destination the outputs are heading to, even though the inputs never stop.

The dominant-term idea

Here is the single most useful idea for limits at infinity: when x is enormous, only the highest-power term matters.

Go back to f(x) = \dfrac{3x^2 + 5}{x^2 + 1}. When x = 1000, the numerator is 3(1{,}000{,}000) + 5 = 3{,}000{,}005. The +5 contributes almost nothing — it is five parts in three million. The denominator is 1{,}000{,}001. The +1 contributes almost nothing. The function is essentially \frac{3x^2}{x^2} = 3, with a tiny correction that fades as x grows.

This is the dominant-term principle: in a polynomial, the term with the highest power of x dominates all others when x is large. The lower-order terms become negligible — not exactly zero, but as close to zero as you want, provided you take x large enough.

To make this rigorous, divide every term in both the numerator and the denominator by the highest power of x that appears:

As x \to \infty, \frac{5}{x^2} \to 0 and \frac{1}{x^2} \to 0. So the expression tends to \frac{3 + 0}{1 + 0} = 3.

That is the technique. Divide top and bottom by the right power of x, watch the small terms die, read off the answer.

Why does this work? Because for any constant k and any positive integer n, \frac{k}{x^n} \to 0 as x \to \infty. This is the fundamental fact behind all limits at infinity: any constant divided by a growing power of x is eventually negligible. It follows from the Archimedean property of real numbers — for any positive number \varepsilon, you can find an x large enough that \frac{k}{x^n} < \varepsilon.

Here is a second example to make the dominant-term idea feel automatic. Take f(x) = \frac{x + 7}{2x - 3}. Divide top and bottom by x:

The answer is \frac{1}{2}, which is just the ratio of the leading coefficients (1 and 2). The lower-order terms (+7 in the numerator, -3 in the denominator) are completely irrelevant once x is large.

The three cases for rational functions

A rational function has the form \dfrac{P(x)}{Q(x)} where P and Q are polynomials. Let \deg P be the degree of the numerator and \deg Q the degree of the denominator. The limit at infinity depends on the comparison of these two degrees.

Why these three cases work. Divide top and bottom by x^m (where m = \deg Q). In Case 1, the numerator's terms all have negative powers of x after division, so they all tend to 0. In Case 2, the numerator and denominator each have exactly one constant term (their leading coefficients), and everything else tends to 0. In Case 3, the numerator still has positive powers of x left over, so it grows without bound while the denominator settles to a constant.

Horizontal asymptotes

When the limit at infinity exists and is a finite number L, the line y = L is a horizontal asymptote of the graph.

A function can have at most two horizontal asymptotes: one as x \to \infty and one as x \to -\infty. Some functions have the same asymptote in both directions (like \frac{3x^2+5}{x^2+1}, which tends to 3 in both directions). Others have different asymptotes in different directions (like \frac{x}{|x| + 1}, which tends to 1 as x \to +\infty and -1 as x \to -\infty). And many functions — like f(x) = x^2 — have no horizontal asymptote at all, because the outputs grow without bound.

A horizontal asymptote is not a barrier. The function can cross its asymptote — the asymptote only describes the long-run behaviour, not the behaviour at any specific x value. For instance, f(x) = \frac{\sin x}{x} has the asymptote y = 0, but the function crosses y = 0 infinitely many times.

Limits involving radicals

When a limit at infinity involves a square root, the dominant-term trick still works, but you have to be careful about signs.

Consider \lim_{x \to \infty}\left(\sqrt{x^2 + 3x} - x\right). If you try to evaluate this directly, you get \infty - \infty — which is indeterminate. The two terms both grow without bound, and the difference could be anything.

The standard trick is rationalisation: multiply and divide by the conjugate.

Now divide numerator and denominator by x (which is positive since x \to +\infty):

As x \to \infty, \frac{3}{x} \to 0, so \sqrt{1 + \frac{3}{x}} \to 1, and the expression tends to \frac{3}{1 + 1} = \frac{3}{2}.

The sign of x matters. When x \to +\infty, \sqrt{x^2} = x (positive). When x \to -\infty, \sqrt{x^2} = |x| = -x (because x is negative, |x| = -x). Forgetting this sign flip is one of the most common mistakes.

To see the sign issue in action, try \lim_{x \to -\infty}\left(\sqrt{x^2 + 3x} + x\right). Notice the +x, not -x. When x is a large negative number, \sqrt{x^2 + 3x} \approx |x| = -x, so the expression is approximately -x + x = 0... but not exactly. Rationalise by multiplying by \frac{\sqrt{x^2 + 3x} - x}{\sqrt{x^2 + 3x} - x} to get \frac{x^2 + 3x - x^2}{\sqrt{x^2 + 3x} - x} = \frac{3x}{\sqrt{x^2 + 3x} - x}. Now divide by |x| = -x (since x < 0): \frac{-3}{-\sqrt{1 + 3/x} - 1} = \frac{-3}{-\sqrt{1 + 3/x} - 1}. As x \to -\infty, 3/x \to 0, and the expression tends to \frac{-3}{-1 - 1} = \frac{-3}{-2} = \frac{3}{2}. The limit is \frac{3}{2} — the same value as the earlier example, despite the expression looking different. The sign bookkeeping is where the difficulty lies.

Worked examples

Common confusions

• "\infty - \infty = 0." No. The form \infty - \infty is indeterminate. \sqrt{x^2 + 3x} - x \to \frac{3}{2}, not 0. x^2 - x \to \infty. (x+1) - x = 1 for all x. The same "form" gives three completely different answers.

• "\frac{\infty}{\infty} = 1." Also indeterminate. \frac{2x}{x} = 2 for all x, so the limit is 2, not 1. \frac{x^2}{x} = x \to \infty. \frac{x}{x^2} = \frac{1}{x} \to 0. The ratio depends on the rates of growth.

• "\sqrt{x^2} = x." Only when x \geq 0. In general, \sqrt{x^2} = |x|. When x \to -\infty, \sqrt{x^2} = -x (positive, because x is negative). This sign error wrecks many limits at -\infty.

• "A function cannot cross its horizontal asymptote." False. The function f(x) = \frac{\sin x}{x} crosses y = 0 at every x = n\pi (n \neq 0), yet y = 0 is its horizontal asymptote. A horizontal asymptote is a statement about what happens eventually, not a fence.

• "If the degree of the numerator is higher, the limit is +\infty." Not necessarily +\infty. The sign of the limit depends on the signs of the leading coefficients. \frac{-x^3}{x^2} = -x \to -\infty as x \to +\infty.

Going deeper

The formal definition

The intuitive idea "as x grows, f(x) gets close to L" has a precise version.

\lim_{x \to \infty} f(x) = L means: for every \varepsilon > 0 (no matter how small), there exists a number M such that whenever x > M, |f(x) - L| < \varepsilon.

In words: pick any tolerance — say you want f(x) to be within 0.001 of L. Then there exists a threshold M such that once x passes M, f(x) stays within 0.001 of L — and stays there forever. The function does not wander away. Pick a tighter tolerance, like 0.0001, and there is a (possibly larger) threshold that works. No matter how tight the tolerance, a threshold exists.

For f(x) = \frac{1}{x} and L = 0: given \varepsilon > 0, choose M = \frac{1}{\varepsilon}. Then for all x > M, |f(x) - 0| = \frac{1}{x} < \frac{1}{M} = \varepsilon. That completes the proof that \lim_{x \to \infty}\frac{1}{x} = 0. The explicit choice M = 1/\varepsilon is the key move — it translates the abstract tolerance \varepsilon into a concrete threshold.

When the limit is infinity

Sometimes the notation \lim_{x \to \infty} f(x) = \infty is used. This does not mean "the limit exists and equals infinity" — infinity is not a number, and limits are numbers. It means: f(x) grows without bound. For every N (no matter how large), there exists M such that x > M implies f(x) > N. The function eventually exceeds any finite target.

Exponential vs. polynomial growth

When an exponential and a polynomial compete, the exponential always wins. Specifically:

No matter how large the degree n, e^x eventually dwarfs x^n. The exponential function grows faster than any polynomial. The proof uses repeated application of L'Hopital's rule (or direct estimation), and the result is one of the most important asymptotic facts in all of analysis.

Similarly, \ln x grows slower than any positive power of x:

The growth hierarchy is: logarithms < polynomials < exponentials. Every function in a slower class is eventually overtaken by every function in a faster class, no matter the constants involved. Even 1000000\ln x is eventually smaller than x^{0.001}, and even x^{1000000} is eventually smaller than e^x. The constants only delay the inevitable — they never change the final outcome.

To see why \frac{x^n}{e^x} \to 0, here is a sketch for n = 1. For x > 0, e^x = 1 + x + \frac{x^2}{2} + \cdots > \frac{x^2}{2}. So \frac{x}{e^x} < \frac{x}{x^2/2} = \frac{2}{x} \to 0. For higher n, a similar comparison with a higher-degree term in the Taylor expansion of e^x does the job.

Oblique asymptotes

When \deg P = \deg Q + 1 — the numerator is exactly one degree higher — the function does not have a horizontal asymptote, but it does have an oblique (slant) asymptote. Perform polynomial long division: \frac{P(x)}{Q(x)} = (mx + b) + \frac{R(x)}{Q(x)}, where the remainder R has degree less than Q. As x \to \infty, \frac{R(x)}{Q(x)} \to 0, so the function approaches the line y = mx + b. The graph hugs this slanted line at the extremes, just as it would hug a horizontal asymptote.

Take f(x) = \frac{x^2 + 1}{x}. Divide: \frac{x^2 + 1}{x} = x + \frac{1}{x}. As x \to \infty, the remainder \frac{1}{x} \to 0, so the oblique asymptote is y = x. The graph of f looks like the line y = x with a small bump near the origin. Far from the origin, the bump is invisible — the function and the line become indistinguishable.

Limits at infinity for non-rational functions

Not all functions you meet are rational. Some general principles:

• Exponentials dominate polynomials. \lim_{x \to \infty}\frac{e^x}{x^{100}} = \infty. No polynomial, however high its degree, can keep up with e^x.

• Polynomials dominate logarithms. \lim_{x \to \infty}\frac{\ln x}{x^{0.01}} = 0. Even a tiny positive power of x eventually outgrows \ln x.

• The hierarchy. As x \to \infty, functions are ordered by growth rate: \ln x \ll x^a \ll e^x \ll e^{x^2} for any fixed a > 0. The symbol \ll means "grows negligibly compared to." In a limit involving a ratio of two such functions, the faster-growing one wins — it goes to \infty in the numerator, or kills the fraction to 0 in the denominator.

These facts are behind many JEE problems that look hard but have immediate answers once you see the growth hierarchy at work.

Where this leads next

Understanding limits at infinity unlocks the long-run behaviour of functions — the foundation for curve sketching, asymptotic analysis, and much of applied mathematics.

• Exponential and Logarithmic Limits — the standard limits involving e^x and \ln x, which control exponential growth and decay.
• Sandwich Theorem — a different technique for proving limits, especially when the function oscillates.
• Continuity — what it means for a function to have no jumps, and how limits make that idea precise.
• Derivative — the limit that measures instantaneous rate of change, the central object of calculus.
• Rational Functions — the algebraic structure of the functions whose limits at infinity you just analysed.