Lim of 1/x as x Approaches Infinity: How to Evaluate It

The limit of 1/x as x approaches infinity is 0. As x grows without bound, the fraction 1/x shrinks toward zero, because dividing 1 by ever-larger numbers produces ever-smaller results. This is written lim(x→∞) 1/x = 0. The function never actually reaches zero for any finite x, but it gets arbitrarily close, which is exactly what a limit at infinity describes. Finding the lim of 1/x as x approaches infinity is one of the simplest and most foundational limits in calculus, and it underpins results like the horizontal asymptote of rational functions.

Understanding the Expression

A limit at infinity asks what value a function approaches as its input grows larger and larger in the positive direction. It is not asking what happens at infinity — infinity is not a number you can plug in — but what trend the outputs follow as x becomes arbitrarily large.

Here, the function is f(x) = 1/x. Its domain excludes x = 0, but every large positive x is fair game. So the question is purely about behavior on the far right of the number line.

The key intuition is reciprocal growth. When the denominator of a fraction grows while the numerator stays fixed at 1, the fraction’s value falls. Doubling x halves the result. Multiplying x by ten divides the result by ten. There is no floor that stops this decline at some positive number, so the only remaining candidate for the limit is 0.

This same reasoning extends to constants over powers of x: for any fixed positive exponent, c/xⁿ tends to 0 as x → ∞. That generalization appears constantly in calculus, from Khan Academy limit lessons to more advanced analysis.

A Limit at Infinity Versus a Limit at a Point

A limit at a finite point, like lim(x→2) f(x), describes behavior near a specific input. A limit at infinity describes behavior as the input escapes all bounds. The two use different formal definitions, though both share the same “approaches” language.

The Result and Proof

The limit equals 0:

lim(x→∞) 1/x = 0

Two standard tools justify this beyond intuition.

Squeeze Theorem

For all x ≥ 1, we have 0 ≤ 1/x ≤ 1, and as x → ∞ the upper bound shrinks toward 0. Since 1/x is trapped between 0 and a quantity tending to 0, it must tend to 0 as well. To see the squeeze more clearly, consider any fixed positive number a. For x > a, we have 0 ≤ 1/x ≤ 1/a, and 1/a can be made as small as we like by choosing a large enough. That is the essence of the squeeze: the middle term is pinned between a lower bound of 0 and an upper bound that can be driven arbitrarily close to 0. Since 1/x is always positive for positive x, it can never dip below 0, and since the upper bound collapses, it has nowhere to go but 0.

Epsilon-Delta Style Argument

The formal definition of a limit at infinity says: for every ε > 0, there exists a number M such that |1/x − 0| < ε whenever x > M.

Proof sketch: given ε > 0, choose M = 1/ε. Then for any x > M, we get 1/x < 1/M = ε. Since 1/x is positive, |1/x − 0| = 1/x < ε. The condition is satisfied, so the limit is confirmed as 0. This is the standard ε-M formulation; Wolfram MathWorld documents the limit definition in its usual notation. The rigor here matters because it removes all dependence on “seems to approach.”

Why Not a Smaller Positive Number?

Suppose someone claims the limit is 0.5. Then choose ε = 0.1. Once x > 2, the value 1/x is already below 0.5, and it keeps falling. No positive number survives the ε test, which is why 0 is the unique answer.

A Worked Example

Take x = 1,000. Then 1/x = 0.001. Now compare x = 10,000: 1/x = 0.0001. Multiplying the input by ten divided the output by ten, exactly as the reciprocal-growth intuition predicts. Push x to 1,000,000 and 1/x falls to 0.000001. Each step confirms the same pattern — the values march toward 0 from above and never turn negative for positive x. Pick any small positive target, say 0.0001. You can find an x large enough that 1/x is smaller than it — specifically, any x greater than 10,000. That works for every positive target, no matter how tiny. Because you can always push 1/x below any positive threshold, the value 0 is the only possible limit.

Note that x must grow in the positive direction. As x → −∞, the same limit of 0 holds, but the values approach from below (they stay negative). The one-sided behavior differs in sign, not in destination. MIT OpenCourseWare covers this distinction in its single-variable calculus materials.

Visual Interpretation

Plot y = 1/x for positive x. The curve starts high near x = 1 (where y = 1), drops steeply through (2, 0.5) and (10, 0.1), then flattens dramatically. By x = 100 the curve sits at y = 0.01, visually hugging the x-axis.

That flattening reveals a horizontal asymptote: the line y = 0. An asymptote is a line the graph approaches but does not cross (in this region). As x → ∞, the curve gets arbitrarily close to y = 0 without touching it for any finite x.

The graph also shows the behavior is not symmetric in sign across the full domain. For x > 0 the curve stays above the axis; for x < 0 it approaches the same asymptote from below, since 1/x is negative there. The vertical asymptote at x = 0 is a separate feature — it concerns the input approaching zero, not infinity, and is unrelated to the limit we evaluated.

Seeing the asymptote confirms the algebra: the horizontal line y = 0 is the geometric picture of lim(x→∞) 1/x = 0.

Key Takeaways

• The limit of 1/x as x approaches infinity is 0, written lim(x→∞) 1/x = 0.
• As x grows without bound, 1/x shrinks toward zero but never reaches it for any finite x.
• A worked example (1,000 → 0.001, 10,000 → 0.0001) illustrates the decline.
• The result is proved with the squeeze theorem and confirmed by the ε-M definition of a limit at infinity.
• Graphically, y = 0 is a horizontal asymptote of y = 1/x as x → ∞.