Limit laws and one-sided limits

The missing value does not stop the approach

Suppose a calculator evaluates f(x)=(x²−4)/(x−2). At x=2 its denominator is zero, so this formula supplies no value. Yet x=3/2 gives 7/2, x=7/4 gives 15/4, x=9/4 gives 17/4 and x=5/2 gives 9/2. Inputs nearer two suggest outputs nearer four.

The identity x²−4=(x−2)(x+2) explains these values. Whenever x≠2, cancellation gives f(x)=x+2. Therefore the output’s distance from four is exactly the input’s distance from two: |f(x)−4|=|x−2|. This identity works for every allowed nearby input, not just the four records.

Writing lim(x→2) f(x)=4 describes this approach. It does not assign f(2). Leaving f(2) undefined or separately setting f(2)=9 gives the same limit, because the limiting test never uses x=2. The hollow marker on the graph records the omitted point of the approaching line; a separate filled marker records an assigned point value.

The precise promise behind a limit

Prepare by substituting into expressions, plotting coordinates and reading absolute-value inequalities. The linked lessons practise those skills. All limit notation needed here is introduced locally. First read the examples and the one-sided comparison; return to the product and quotient proofs when you can follow the error inequalities.

For a real number u, |u| is its distance from zero; |x−a| is the distance between x and a. We use |u+v|≤|u|+|v| and |uv|=|u||v|. The first inequality holds because −|u|≤u≤|u| and −|v|≤v≤|v|, so their sum lies between −(|u|+|v|) and |u|+|v|. Also |g|≥|M|−|g−M| follows by applying it to M=g+(M−g).

Our two-sided laws concern real-valued functions defined at every input with 0<|x−a|<r for some r>0. This is a punctured interval around a: the middle input is removed. When several functions occur, use an interval where all of them are defined. Functions on sparser domains need a domain-relative definition; we do not silently extend these interval statements to a missing approach side.

The assertion lim(x→a) f(x)=L, with finite real L, means: for every ε>0 there is δ with 0<δ≤r such that every input satisfying 0<|x−a|<δ also satisfies |f(x)−L|<ε. Epsilon, ε, is the requested output tolerance; delta, δ, is a sufficient input radius. The radius may depend on the tolerance. It must control every allowed input inside it, and may be smaller than necessary.

A finite table or a drawn curve helps form a prediction. It cannot check every positive tolerance and every nearby real input. The identities and inequalities below establish those universal promises. None of these laws uses infinite limits or a rule for the symbol 0/0.

One approach has at most one finite limit

Suppose both L and M were limits at the same approach and L≠M. Let d=|L−M|>0 and request tolerance d/3 from each limit. Use the smaller of their positive radii. There is an allowed input within it, for example a plus half that radius in a two-sided punctured interval. At that input, d≤|L−f(x)|+|f(x)−M|<2d/3, a contradiction. Thus L=M. On a specified single side, choose the input on that side instead.

For a constant function f(x)=c, the error |f(x)−c| is zero, so any positive radius inside its domain works for every ε>0. For f(x)=x, choosing δ=min(r,ε) gives |x−a|<ε. Hence lim(x→a)c=c and lim(x→a)x=a.

For f(x)=2x+3 at a=0, the error from L=3 is 2|x|. Choose δ=ε/2, or the minimum of this and a needed domain radius. The shaded band in the diagram is the output requirement 3−ε<y<3+ε. The narrower input strip guarantees it. Boundary lines show strict inequalities: equality at |x|=δ is outside the tested strip.

Add errors and share an input radius

Assume f(x)→L and g(x)→M as x→a, both finite, on a shared punctured interval. Then f+g→L+M and f−g→L−M. For any ε>0, ask each function for error less than ε/2 and use the smaller radius. The triangle inequality gives |(f±g)−(L±M)|≤|f−L|+|g−M|<ε.

For a fixed real number c, cf→cL. If c=0 the result is constant zero. Otherwise request |f−L|<ε/|c|; then |cf−cL|=|c||f−L|<ε. Repeated sums use the same idea with a tolerance budget for every summand.

For f=x+1 and g=2x+3 at zero, their limits are 1 and 3. Their sum is 3x+4 and has limit 4; δ=ε/3 is sufficient directly. The law assumes both input limits exist. Its conclusion cannot be reversed. The sign function s(x) equals −1 for x<0 and 1 for x>0; its point value at zero is irrelevant here. This function and its negative each lack a two-sided limit at zero, as the one-sided comparison below proves, although their sum is constantly zero away from zero.

Products need an error bound and a size bound

Under the same hypotheses, fg→LM. Small input errors alone are not yet a product estimate; first keep one factor bounded. Ask |g−M|<1. Then |g|≤|M|+|g−M|<|M|+1.

The identity fg−LM=g(f−L)+L(g−M) splits the product error. For any ε>0, request |f−L|<ε/[2(|M|+1)] and |g−M|<min(1,ε/[2(|L|+1)]). Both denominators are positive even if a limit is zero. Take the smaller of the two radii and the shared domain radius.

At every tested input, |fg−LM|≤|g||f−L|+|L||g−M|<ε/2+ε/2=ε; if L=0 the second term is zero. This proves the product law, including zero limits. It is a uniform nearby estimate, rather than a multiplication of rounded table values.

For (x+1)(2x+3)=2x²+5x+3, the limit at zero is 3. If |x|<1 then |2x²+5x|≤2|x|²+5|x|≤7|x|, so δ=min(1,ε/7) works. A product can have a limit even when a factor does not: x times the sign of x equals |x| away from zero and tends to zero.

A nonzero denominator limit supplies a guard

Assume g→M with M≠0. Request |g−M|<|M|/2. Then |g|≥|M|−|g−M|>|M|/2>0. Thus g cannot vanish anywhere in a sufficiently small punctured interval. The quotient is actually defined there; this guard follows from the limit hypothesis.

For the reciprocal, |1/g−1/M|=|g−M|/(|g||M|)≤2|g−M|/|M|². Given ε>0, request |g−M|<min(|M|/2,ε|M|²/2), and use a corresponding radius. The displayed estimate proves 1/g→1/M.

If also f→L, the product law applied to f and 1/g gives f/g→L/M. For (x+1)/(2x+3) at zero, M=3, so the limit is 1/3. Within |x|<1 the denominator exceeds 1. Directly, |(x+1)/(2x+3)−1/3|=|x|/[3|2x+3|]≤|x|/3, so δ=min(1,3ε) works.

A denominator may vanish at other inputs farther away; only a sufficiently small punctured interval matters. If M=0, this proof and the quotient rule do not apply. That failure does not itself decide whether the quotient has some other finite limit.

Rewrite where the inputs are allowed

Repeated products show that fⁿ→Lⁿ for every fixed positive integer n. A polynomial is a finite sum of constant multiples of these powers, so its limit at a is its value at a. A rational function has the corresponding substituted limit when its denominator’s value is nonzero. This is a consequence of the proved laws, not an instruction to substitute into any fraction.

If two formulas agree throughout a punctured interval around a, each gives exactly the same limit test there. They may have different values at a, or one may lack that value. In our opening example cancellation is justified for x≠2, so (x²−4)/(x−2)=x+2 there and the limit is 4. It does not make the original denominator nonzero at two.

Here is a square-root example with its needed limit justified locally. The real square root √x is the nonnegative number whose square is x, defined for x≥0. Near one, |√x−1|=|x−1|/(√x+1)≤|x−1|. Restrict |x−1|<1 so x>0, then choose δ=min(1,ε). This proves √x→1 as x→1.

For x≥0 and x≠1, multiply (√x−1)/(x−1) by (√x+1)/(√x+1). Since √x+1>0, this is permitted and gives 1/(√x+1). The denominator tends to 2, so the quotient law gives limit 1/2. The rationalisation changes the nearby expression, not the excluded value at one.

Compare the two approaches before joining them

The notation x→a− means approach from inputs x<a, not from negative inputs in general. Require f to be defined throughout some interval (a−r,a). A left-hand limit L means that for each ε>0 some δ>0 makes 0<a−x<δ imply |f(x)−L|<ε. For x→a+, use (a,a+r) and 0<x−a<δ. Shrink δ to stay inside the appropriate domain interval.

For a function defined on a punctured interval on both sides, a two-sided finite limit L exists if and only if both one-sided limits exist and equal L. A two-sided radius already works on each side. Conversely, request a radius on each side for the same ε and L, and take their minimum. Every nearby x≠a lies on one of the two sides, so its error is controlled.

If the one-sided limits differ, uniqueness rules out a common two-sided limit. For s(x)=−1 when x<0 and s(x)=1 when x>0, the left limit at zero is −1 and the right limit is 1. Each side is constant, hence has zero error from its own limit. Assigning s(0)=0 does not bridge the jump.

For h(x)=x+3 on x<0 and h(x)=2x+3 on x>0, both side limits at zero are 3. Given ε>0, δ=min(ε,ε/2) controls both linear errors; thus h→3. Setting h(0)=7 changes the isolated filled marker, not this result.

All the finite algebra laws proved above also hold on a specified single side when every function has the stated limit on that same side. Their proofs use the same error bounds and a minimum radius; only the allowed input inequality changes. Mixing f’s left-hand limit with g’s right-hand limit is not a valid application.

Missing sides and zero denominators are different issues

For √x on its real domain x≥0, the right-hand limit at zero is 0: for every ε>0, choose δ=ε²; then 0<x<δ implies 0<√x<ε. There are no real-domain inputs approaching zero from the left. Under our elementary interval convention this has a right-hand limit, rather than a two-sided limit. A broader limit taken relative to the domain also equals zero; it should be labelled as such.

Compare three quotients at zero, all restricted to x≠0. For x/x=1 the limit is 1. For x²/x=x it is 0. For x/x²=1/x no finite two-sided limit exists. All have numerator and denominator tending to zero. Thus those two limits alone do not determine a quotient’s limit, and 0/0 is not an answer.

To justify the last failure without relying on large plotted values, suppose 1/x had any finite limit L even from the right. Use tolerance 1. For any proposed δ>0, choose a positive x smaller than both δ and 1/(|L|+2). Then 1/x>|L|+2 and |1/x−L|≥1/x−|L|>2, contradicting the required error less than 1. This argument covers every candidate L.

Exact tables can expose the difference between these examples. A finite reciprocal plot is necessarily only a window on the growth. The explorer therefore shows unconnected exact records for these three quotients; it never draws a segment through the excluded input zero or claims that the last row establishes a limit.

Trap the middle function between two limits

Suppose u(x)≤h(x)≤v(x) at every input in a common punctured interval, and u→L and v→L with finite L. For any ε>0, choose a shared radius making L−ε<u and v<L+ε. Then L−ε<h<L+ε, so |h−L|<ε. This proves the squeeze principle; equality with an outer function is allowed.

A useful special case is h(x)=x b(x) near zero with |b(x)|≤C for a fixed C≥0. Then |h(x)|≤C|x|. If C=0, h is zero there. If C>0, take δ=min(r,ε/C); the error is less than ε. The bounded factor need not itself have a limit.

For b=s, the sign function, h=|x|; for b=−s, h=−|x|; for b=s/2, h=|x|/2. Each lies between −|x| and |x| and tends to zero, although these three choices of b all have distinct left and right limits. The envelopes use short dashes; the middle function uses a solid left branch and longer dashes on the right. These patterns distinguish their roles without colour alone.

The inequalities must hold throughout a sufficiently small interval, not just at tested points. Bounds with different limiting values do not force a unique middle limit: −1≤s(x)≤1 is true, but the sign function still jumps at zero.

Predict a result, then check the reason

Choose a case below and predict its finite limit, whether a two-sided limit exists, or which radius is sufficient. The input formula and chosen case remain visible before reveal. Revealing adds exact nearby records, limiting markers and the relevant error or domain guard.

Use the table as a check on arithmetic and the diagram as a check on approach sides. Use the section’s proof to establish the claim for all nearby real inputs. The plotted coordinate scales are numeric; sampled drawings do not replace the exact error estimates.

Before applying a law, name the approach, establish each required finite input limit, share a domain interval, and check the denominator condition. If a condition fails, rewrite locally or prove another bound instead of carrying out an invalid substitution.

Worked examples

Try it yourself

1. Prove lim(x→−2)(3x+5)=−1 by giving a radius for each ε>0.

   Show the answer and reasoning

   Choose δ=ε/3.

   The output error is |3x+5−(−1)|=3|x+2|. Whenever 0<|x+2|<δ, this is less than ε.

2. Find lim(x→2)(3x²−2x+5), stating why substitution is allowed.

   Show the answer and reasoning

   The limit is 13.

   Constants and x have finite limits; repeated product, scalar and sum laws give 3·2²−2·2+5=13. The polynomial is defined on the whole interval.

3. Find lim(x→1)(x²+2)/(x+3) and state the denominator condition.

   Show the answer and reasoning

   The limit is 3/4; the denominator limit is 4≠0.

   The polynomial numerator tends to 3 and denominator to 4. The nonzero limit ensures a small punctured interval with nonzero denominator, as required for division.

4. Both x² and x approach zero as x→0. Does that alone determine the limit of x²/x? Calculate it using a valid argument.

   Show the answer and reasoning

   Those two limits alone do not determine a quotient; this particular quotient tends to 0.

   For x≠0, x²/x=x, so local equality and the identity limit apply. The quotient rule with denominator limit zero would be invalid.

5. A function is 5+x for x<2 and 7+3(x−2) for x>2. What are its limits at two, and can its value at two matter?

   Show the answer and reasoning

   Both one-sided limits and the two-sided limit are 7; its value at two does not matter.

   On the left, |(5+x)−7|=|x−2|; on the right the error is 3|x−2|. The common radius ε/3 works for both sides.

6. A function is x+1 for x<2 and x+4 for x>2. What happens at two?

   Show the answer and reasoning

   The left-hand limit is 3, the right-hand limit is 6, and no finite two-sided limit exists.

   Each linear branch has its stated limit. Their distinct values violate the common-value criterion; an assigned point value cannot help.

7. May you add a left-hand limit 2 of f to a right-hand limit 3 of g and infer a two-sided limit 5 of f+g?

   Show the answer and reasoning

   No. The limits must concern the same approach and the needed functions must be defined there.

   Let f=2 on the left and 20 on the right, and g=30 on the left and 3 on the right. The given side limits hold, but f+g has left limit 32 and right limit 23.

8. For all 0<|x|<1, a function b satisfies |b(x)|≤5. Prove that x b(x)→0 without assuming a limit for b.

   Show the answer and reasoning

   For ε>0 take δ=min(1,ε/5).

   The error |x b(x)| is at most 5|x|, hence less than ε at every tested input. Boundedness is sufficient; a separate limit of b is unnecessary.

9. Does −2≤q(x)≤2 near zero prove that q has a limit there?

   Show the answer and reasoning

   No; for example the sign function lies in those bounds and has no two-sided limit at zero.

   The two constant outer functions have different limits. The squeeze principle requires the same limiting value.

10. For √x on x≥0 at zero, state the right-hand conclusion, a sufficient radius, and the elementary two-sided convention.

    Show the answer and reasoning

    The right-hand limit is 0 with δ=ε². A real left-hand approach is absent, so this is not an elementary two-sided limit.

    For 0<x<ε², √x<ε. A domain-relative limit also equals zero, but it must not be mistaken for the availability of negative-domain inputs.

11. If f+g→0, must f and g each have finite two-sided limits?

    Show the answer and reasoning

    No.

    Take f to be the sign function and g=−f for nonzero inputs. Their sum is constantly zero, while each has different one-sided limits. The sum law is a sufficient rule with hypotheses, not a converse.

What to carry forward

• A finite limit controls every nearby allowed input for every positive output tolerance; the point value is excluded.
• Sum, difference, scalar and product laws require the stated finite limits on the same approach and a shared nearby domain.
• A nonzero denominator limit ensures eventual nonvanishing and permits the quotient law; denominator limit zero requires another argument.
• Equality on a punctured interval preserves a limit even if point values differ.
• With both sides available, a two-sided finite limit exists exactly when the two finite one-sided limits agree.
• Squeeze uses inequalities throughout a neighbourhood and equal outer limits. A bounded factor can still lack a limit.
• Exact records and graphs illustrate claims; identities, inequalities and domain guards establish them.