Infinite limits and limits at infinity

A nearby input and a far-away input ask different questions

Consider f(x)=1/(x−2). Inputs 2+1/2, 2+1/4 and 2+1/8 give outputs 2,4 and 8. Here the input approaches the finite number 2 while the output grows. Input 2 itself is excluded: none of those divisions uses a zero denominator.

Now consider g(x)=2+1/x. Inputs 2,4 and 8 give outputs 5/2,9/4 and 17/8. The inputs grow while the outputs approach the finite number 2. The same symbol ∞ can describe an input direction or an output conclusion; read its position in the limit statement.

In lim(x→2⁺)f(x)=+∞, x approaches 2 from above and f(x) exceeds every fixed positive level eventually. In lim(x→+∞)g(x)=2, x exceeds every fixed input threshold eventually and g(x) approaches 2. The superscript + in 2⁺ names a side, not the sign of the output.

Use the published preparation links for fractions, domains, signed powers and rational graph features. Limits, thresholds and the asymptote tests needed here are defined locally. On a first reading, follow the reciprocal examples and their pictures, then return to the quantified definitions and the full degree proof once the two questions are clear.

Growing without bound means every sufficiently close input

Let f:D→ℝ be a real function. Before writing a right-hand limit at a, require permitted inputs arbitrarily close to a with x>a; for a left-hand limit require them with x<a. Such a is an accumulation point from that side. Without this availability condition, an empty interval could make a purported limit condition vacuously true.

The statement lim(x→a⁺)f(x)=+∞ means: for every real M>0, there exists δ>0 such that every x∈D with 0<x−a<δ satisfies f(x)>M. Replace the conclusion by f(x)<−M for −∞. From the left, use 0<a−x<δ. A two-sided statement uses 0<|x−a|<δ, with domain accumulation from both sides under our two-sided convention.

For f(x)=1/(x−2), choose δ=1/M. If 0<x−2<δ, then 1/(x−2)>M. If 0<2−x<δ, then 1/(x−2)=−1/(2−x)<−M. The two one-sided infinite signs differ, so there is no two-sided +∞ or −∞ conclusion and no finite two-sided limit.

For comparison, a finite domain-relative limit L at an accumulation point a of D requires: for every ε>0, some δ>0 ensures |f(x)−L|<ε for every permitted x with 0<|x−a|<δ. The right and left versions restrict those inputs to their available side. L is a real output, and f(a) is again excluded.

An infinite limit is signed unbounded behaviour, not convergence to a real output called infinity. Every individual permitted f(x) remains a real number. The value f(a), if defined separately, is excluded from these tests. A few large outputs do not prove the condition: it must hold for all permitted inputs sufficiently close on the stated side.

Denominator signs and available sides determine a pole

For 1/(x−2)², the denominator is positive on both available sides. Given M>0, choose δ=1/√M. Then 0<|x−2|<δ implies 1/(x−2)²>M. Thus both one-sided limits and the two-sided infinite limit are +∞. Multiplying by −1 changes all three conclusions to −∞.

More generally, for a positive integer k and nonzero real c, c/(x−a)ᵏ has magnitude |c|/|x−a|ᵏ. Choose δ=(|c|/M)^(1/k). If k is even, both side signs equal the sign of c. If k is odd, the right side has that sign and the left side has the opposite sign. The original domain excludes a throughout.

The real rule h(x)=1/√x has domain x>0. For every M>0, 0<x<1/M² implies h(x)>M. This proves lim(x→0⁺)h(x)=+∞. There are no real domain inputs to the left of 0, so this function has no left-hand limit there under our accumulation convention. Its domain-relative approach is the right-hand approach.

A rational numerator need not stay positive near an excluded input. Factor the original expression, cancel only where permitted and check the sign of each remaining factor on each side. Denominator zero alone does not decide the limit or even establish a pole.

A vertical asymptote requires an infinite one-sided limit

We call x=a a vertical asymptote when at least one available one-sided limit there is +∞ or −∞. Opposite signs are allowed: 1/(x−2) has a vertical asymptote x=2 even though its two-sided limit has neither infinite sign. The rule 1/√x has vertical asymptote x=0 using its available right side.

For r(x)=(x²−4)/(x−2), factor x²−4=(x−2)(x+2). On its original domain x≠2, r(x)=x+2. As x approaches 2, |r(x)−4|=|x−2| becomes arbitrarily small. The missing point is a hole at (2,4); there is no vertical asymptote at 2.

Changing a single value does not change a punctured limit. If a new function equals 1/(x−2) for x≠2 and is assigned the value 7 at x=2, it still has the same infinite one-sided limits and vertical asymptote. Its isolated point (2,7) lies on that line. Do not turn the asymptote definition into a claim that no graph point can ever lie on it.

A plotted branch may leave the top or bottom of a finite window near a pole. The cutoff is a display boundary, not the function stopping or reaching infinity. Separate branches across excluded inputs, and mark a finite hole with an open marker rather than joining it into the original graph.

Far-away inputs can approach a finite output

For a limit as x→+∞, require D to be unbounded above: for every real threshold there is a permitted input above it. For x→−∞ require D unbounded below. The input is never equal to ∞ or −∞; these symbols name directions through the real domain.

We write lim(x→+∞)f(x)=L, with L real, when for every ε>0 there is N>0 such that every x∈D with x>N satisfies |f(x)−L|<ε. For x→−∞ use x<−N. The positive number ε is an allowed output error; N is an input threshold, and its choice may depend on ε.

For g(x)=2+1/x, choose N=1/ε. If x>N or x<−N, then |g(x)−2|=1/|x|<ε. This proves a finite limit of 2 at each end. The equality |g(N)−2|=ε is not a failure: the definition uses the strict input inequality x>N.

For an integer k≥1, 1/xᵏ→0 at each end because |1/xᵏ|=1/|x|ᵏ<ε whenever |x|>ε^(−1/k). Its sign may differ at the two ends for odd k, but both finite limits are zero. A function on a bounded domain has no limit at +∞ or −∞ under the availability convention.

An end asymptote is a statement about a vanishing difference

A horizontal asymptote y=L means f(x)→L at at least one available end. Different ends can have different horizontal asymptotes. The graph may meet or cross such a line at finite inputs; the end limit says only that the output difference eventually becomes small.

For f(x)=2+(x−1)/(x²+1), f(1)=2. Its difference from 2 changes sign there because x²+1 is positive. For |x|≥1, |f(x)−2|≤(|x|+1)/x²≤2/|x|, which tends to zero. Thus y=2 is crossed at (1,2) and remains a horizontal asymptote at both ends.

A slant asymptote y=mx+b with m≠0 means f(x)−(mx+b)→0 at an available end. A quotient approaching 1, such as f(x)/(mx+b)→1, alone is weaker: f(x)=x+1 and the line y=x have ratio tending to 1 but their difference is always 1.

Division gives (x²+1)/(x−1)=x+1+2/(x−1) for x≠1. For |x|>2, |x−1|≥|x|−1>|x|/2, so the remainder magnitude is less than 4/|x| and tends to zero. Hence y=x+1 is its slant asymptote at both ends. The vertical exclusion x=1 and the end comparison answer different questions.

Far-away inputs can also force signed unbounded outputs

We write lim(x→+∞)f(x)=+∞ when for every M>0 some N>0 makes f(x)>M for every permitted x>N. For −∞ output replace this by f(x)<−M. At the negative input end use x<−N. Keep the domain unbounded in the chosen direction.

For x², choosing N=√M makes x²>M at either end, so both end limits are +∞. For x³, choosing N=M^(1/3) gives x³>M when x>N and x³<−M when x<−N. Thus its positive-end limit is +∞ and its negative-end limit is −∞.

Unbounded magnitude alone is insufficient for a signed infinite limit. On the positive-integer domain, define f(n)=n for even n and f(n)=−n for odd n. Its magnitude grows without bound, but every tail contains both large positive and large negative outputs. It has neither a finite end limit nor +∞ nor −∞ as n→+∞.

The symbol ∞ is not a real number available for arithmetic. For example, x−x=0, 2x−x=x and x−2x=−x have different limits at +∞ although each separate positive term tends to +∞. Likewise 1/x² and x² both tend to +∞ at different approaches, without licensing a substitution of infinity into a formula.

Prove leading-term dominance rather than just dropping terms

Let p(x)=aₙxⁿ+⋯+a₁x+a₀ with n≥1 and aₙ≠0. For x≠0, factor p(x)=xⁿ(aₙ+aₙ₋₁/x+⋯+a₀/xⁿ). Every term inside the parentheses other than aₙ tends to zero as |x| grows.

Here is an explicit bound. Put C=|a₀|+⋯+|aₙ₋₁|. For |x|≥1, the magnitude of the parenthesis's error is at most C/|x|. Once |x|>max(1,2C/|aₙ|), its sign is the sign of aₙ and its magnitude is at least |aₙ|/2. Therefore |p(x)|≥(|aₙ|/2)|x|ⁿ, which exceeds every fixed M eventually.

At +∞ the polynomial has the sign of aₙ. At −∞ multiply that sign by (−1)ⁿ: even powers keep it, odd powers reverse it. The inequality proves the unbounded conclusion as well as the eventual sign. A nonzero constant polynomial has its constant finite limit, and the zero polynomial has limit zero; neither needs a fictitious degree for zero.

Rational end behaviour follows from a nonzero normalised denominator

Let p and q be nonzero polynomials of degrees n and m with leading coefficients aₙ and bₘ, and q not the zero polynomial. On the original real domain q(x)≠0, factor p(x)/q(x)=x^(n−m)·A(x)/B(x), where A(x)=aₙ+aₙ₋₁/x+⋯ and B(x)=bₘ+bₘ₋₁/x+⋯. This identity uses x≠0, harmless in sufficiently far tails.

The previous error bound gives A→aₙ and B→bₘ. Eventually |B|≥|bₘ|/2, so the denominator is nonzero throughout both far tails. To justify the quotient, write A/B−aₙ/bₘ=[bₘ(A−aₙ)−aₙ(B−bₘ)]/(bₘB). Its magnitude is at most 2|A−aₙ|/|bₘ|+2|aₙ||B−bₘ|/|bₘ|², which tends to zero. Thus A/B→c=aₙ/bₘ≠0.

If n<m, the factor x^(n−m)=1/x^(m−n) tends to zero and A/B is eventually bounded, so the quotient tends to zero at both ends. If n=m, the limit is c at both ends. If n>m, A/B eventually has sign c and magnitude at least |c|/2; multiplication by x^(n−m) then gives signed infinite limits. At −∞ reverse the sign exactly when n−m is odd.

These are end rules, not a rule for finite denominator zeros. They prove horizontal asymptotes when n≤m; when n>m there is no finite horizontal asymptote. A degree difference of one suggests a slant line, whose exact remainder must still tend to zero after polynomial division. An identically zero numerator gives zero on its original domain and zero end limits whenever that domain is unbounded.

At the negative end, √(x²) equals |x|

For real x, √(x²)=|x|, not always x. Hence √(x²)/x equals 1 for x>0 and −1 for x<0. Its end limits are different even though its numerator and denominator both have large magnitudes.

For h(x)=√(x²+1)/x on x≠0, rewrite h(x)=(|x|/x)√(1+1/x²). If u≥0, rationalising gives 0≤√(1+u)−1=u/[√(1+u)+1]≤u/2. With u=1/x², the error tends to zero. Consequently h→1 at +∞ and h→−1 at −∞.

Keep the principal square root and its real-domain condition before cancelling or dividing. Factoring a square under a root introduces |x|; changing it to x at the negative end can reverse a correct answer. No differentiation or rule for an infinity-over-infinity symbol is needed here.

Use exact tables to test signs, then return to the full inequalities

The explorer separates nearby-input cases, end cases and strict-threshold examples. Predict the two signs or limits before revealing the exact output table. Its rational values are exact; smooth plotted curves use finite sampling on explicitly labelled linear axes.

A table is finite evidence. A pole can leave the drawing window; an end limit can be invisible until an input exceeds the current window. The sign and threshold proofs cover all sufficiently close or far permitted inputs, not only the displayed rows. Infinity is never a plotted endpoint or an accepted formula input.

Return to the definitions by naming the challenge M or ε, choosing a suitable δ or N and verifying the inequality for every input in the required domain region. Then revisit the leading-term proof, the nonzero denominator bound and the absolute-value step. These three checks prevent most sign, degree and cancellation errors.

Worked examples

Try it yourself

1. In lim(x→3⁻)f(x)=+∞, does x go to negative infinity?

   Show the answer and reasoning

   No. It approaches the finite number 3 from below.

   The minus superscript names the input side. The positive infinity on the right describes the output growing beyond every positive level.

2. Find both one-sided limits of −2/(x−4)² at 4.

   Show the answer and reasoning

   Both are −∞; x=4 is a vertical asymptote.

   The squared denominator is positive. Choosing δ=√(2/M) makes the negative output less than −M on each side.

3. Find the left and right limits of 5/(x+2)³ at −2.

   Show the answer and reasoning

   Left −∞; right +∞.

   The odd denominator changes sign. With δ=(5/M)^(1/3), every permitted input within that distance has magnitude exceeding M.

4. What left-hand limit at 0 does 1/√x have on its real domain?

   Show the answer and reasoning

   None under the side-accumulation convention.

   Its domain is x>0, so no negative permitted input approaches 0. The right-hand limit is +∞, proved by 0<x<1/M².

5. Does denominator zero at 2 prove that (x²−4)/(x−2) has a vertical asymptote?

   Show the answer and reasoning

   No. It has a finite hole at (2,4).

   On x≠2 it equals x+2; |(x+2)−4|=|x−2| tends to zero rather than growing without bound.

6. For 2+1/x and ε=1/4, is x=4 included in the guarantee x>N when N=4?

   Show the answer and reasoning

   No. At x=4 the error is exactly 1/4.

   The strict conclusion error<ε is guaranteed for x>4, or at the other end x<−4. A boundary equality is kept outside that stated guarantee.

7. Find both end limits of (2x+1)/(x³+2).

   Show the answer and reasoning

   Both are 0.

   The degree difference is −2. Normalising by x³ gives (2/x²+1/x³)/(1+2/x³); the denominator tends to 1 and is eventually nonzero.

8. Find both end limits of (−4x²+7)/(2x²+3).

   Show the answer and reasoning

   Both are −2.

   Divide by x². The numerator tends to −4 and denominator to 2, giving the finite leading-coefficient ratio −2.

9. Find both end limits of −x⁴/(x²+1).

   Show the answer and reasoning

   Both are −∞.

   The leading ratio is −1 and the degree difference is even, equal to 2. The negative sign persists at both ends, with magnitude growing without bound.

10. Find the positive and negative end limits of (−2x³+1)/(x²+1).

    Show the answer and reasoning

    At +∞: −∞. At −∞: +∞.

    The leading ratio is −2 and the degree difference is odd. The negative input end reverses the leading sign.

11. What are the end limits of the constant rules 7 and 0?

    Show the answer and reasoning

    They are 7 and 0 respectively at either end of an unbounded real domain.

    Their errors from the stated values are identically zero. The zero polynomial is handled directly without a degree assignment.

12. Can 2+(x−1)/(x²+1) cross its horizontal asymptote?

    Show the answer and reasoning

    Yes, at (1,2); the asymptote is y=2 at both ends.

    The difference from 2 vanishes at 1 and changes sign there, while its tail magnitude is bounded by 2/|x| for |x|≥1.

13. Does (x+1)/x→1 prove y=x is a slant asymptote of f(x)=x+1?

    Show the answer and reasoning

    No. The difference f(x)−x equals 1, not a quantity tending to zero.

    A relative ratio can approach 1 while an absolute vertical difference stays nonzero. The slant-asymptote definition tests the difference.

14. Find the two end limits of √(x²)/x on x≠0.

    Show the answer and reasoning

    At +∞: 1. At −∞: −1.

    The principal root is |x|. The quotient is exactly 1 on positive inputs and −1 on negative inputs.

15. Can a function with declared domain 0<x<1 have a limit at +∞ in this lesson's convention?

    Show the answer and reasoning

    No.

    The domain is not unbounded above. There are no permitted inputs beyond a large threshold, so such a limit is not defined by a vacuous implication.

16. On positive integers, f(n)=n for even n and −n for odd n. Does |f(n)|→+∞ imply f(n)→+∞?

    Show the answer and reasoning

    No; f has neither signed infinite end limit.

    Every tail includes arbitrarily large negative odd outputs and positive even outputs. Either proposed signed inequality fails on infinitely many permitted inputs.

17. Find the limits at +∞ of x−x, 2x−x and x−2x.

    Show the answer and reasoning

    They are 0, +∞ and −∞ respectively.

    Simplify each real expression before taking the limit. The informal form infinity minus infinity does not determine an answer.

18. If 1/(x−2) on x≠2 is extended by assigning f(2)=7, does x=2 stop being a vertical asymptote?

    Show the answer and reasoning

    No.

    The one-sided punctured limits exclude x=2 and are unchanged. The isolated assigned point can lie on the vertical asymptote line.

What to carry forward

• Distinguish the input approach from the signed or finite output conclusion.
• Require domain inputs approaching the named side or extending unboundedly in the named end direction.
• An infinite limit must hold beyond every level for all sufficiently close or far permitted inputs; unbounded magnitude alone is insufficient.
• A hole has a finite approached output; a vertical asymptote needs an infinite limit from at least one available side.
• Horizontal and slant end asymptotes test a difference tending to zero, and horizontal asymptotes can be crossed.
• Rational end rules require a nonzero normalised denominator, leading coefficients and the parity of the degree difference.
• Retain √(x²)=|x| and the original domain. Infinity is never substituted as an ordinary real input or output.