Limits and Continuity
Understand nearby behavior, calculate limits, distinguish holes from asymptotes, and make piecewise models continuous.
Why study nearby behavior?
A model may be undefined at one input even though its nearby outputs follow a clear pattern. Limits describe that pattern. This lets us study a missing value, a sudden change, or a long-run trend before asking how quickly a quantity changes.
Learning goals
Read limits from formulas, tables and graphs; justify algebraic transformations; distinguish one-sided, infinite and end-behavior limits; and test continuity using all three conditions.
A limit describes nearby outputs
Finite limit
We write \(\lim_{x\to c}f(x)=L\) when the values of \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to, but different from, \(c\). For the two-sided convention used here, nearby inputs are available on both sides.
The value at the destination is a separate question. Follow the nearby outputs first; do not let a single filled point on a graph determine the limit.
A missing output
For \(f(x)=(x^2-9)/(x-3)\), what happens as \(x\to3\)?
Show worked solution
For \(x\ne3\), factor and cancel: \(f(x)=x+3\). Nearby outputs approach 6, so the limit is 6, although the original function is undefined at 3.
| x | 2.9 | 2.99 | 3.01 | 3.1 |
|---|---|---|---|---|
| f(x) | 5.9 | 5.99 | 6.01 | 6.1 |
Keep in mind
A table suggests a limit but does not prove it. A finite collection of sampled values can miss oscillation or a narrow change.
Move the filled point
Define \(g(x)=x+3\) for \(x\ne3\) and \(g(3)=20\). Find the limit at 3 and compare it with the function value.
Hint
Only nearby values contribute to the limit.
Show worked solution
The limit is 6. The function value is 20. Changing one output does not change the nearby rule.
A precise closeness condition
The finite limit is L if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that every allowed input with \(0<|x-c|<\delta\) satisfies \(|f(x)-L|<\varepsilon\).
Think of ε as an output tolerance. We must find an input tolerance that works for every sufficiently nearby input, not just for the values a table happens to sample.
Control the error
Show that \(3x+1\to7\) as x→2.
Show worked solution
The output error is \(|3x+1-7|=3|x-2|\). Given any ε>0, choose δ=ε/3. Then 0<|x−2|<δ makes the output error smaller than ε. This establishes every requested tolerance, not one numerical example.
Approach a missing point
A limit is not the assigned value
One-sided limits and jumps
One-sided limits
\(\lim_{x\to c^-}f(x)\) uses inputs less than \(c\); \(\lim_{x\to c^+}f(x)\) uses inputs greater than \(c\). A finite two-sided limit exists exactly when both one-sided limits exist and equal the same finite number.
Piecewise models require us to follow the correct formula on each side. The inequality containing the endpoint determines its value, not either neighboring limit.
Matching two formulas
Let \(f(x)=2x\) for \(x<3\) and \(f(x)=(x^2-9)/(x-3)\) for \(x>3\). Find the limit at 3.
Show worked solution
From the left, 2x approaches 6. From the right, cancellation gives x+3, also approaching 6. Therefore the two-sided limit is 6; no value at 3 is required.
A delivery fee jumps
A delivery fee is ₱50 for a weight below 2 kg and ₱80 for a weight at least 2 kg. Find both one-sided limits at 2 kg. Does a two-sided limit exist?
Hint
Compare the constant fees on opposite sides.
Show worked solution
The left limit is ₱50 and the right limit is ₱80. They differ, so no two-sided limit exists. The fee at exactly 2 kg is ₱80.
Limit laws and direct substitution
Limit laws
When \(f(x)\to A\) and \(g(x)\to B\) are finite, sums, differences and products approach the corresponding combinations of \(A\) and \(B\). Quotients approach \(A/B\) provided \(B\ne0\). Continuous powers and roots may be applied on their real domains.
These laws explain why substitution works for polynomials and for rational functions with a nonzero denominator at the target. Check the denominator before dividing.
Substitution with a condition
Evaluate \(\lim_{x\to2}(x^3-2x^2+7x-1)\).
Show worked solution
A polynomial is continuous, so substitute: \(8-8+14-1=13\).
A quotient limit
Evaluate \(\lim_{x\to-3}((x+1)/(2x+3))^3\).
Hint
Verify the denominator is nonzero before substitution.
Show worked solution
The denominator tends to −3. The quotient tends to \((-2)/(-3)=2/3\), so the cube tends to \(8/27\).
When substitution gives zero over zero
Indeterminate form
\(0/0\) is not a number and is not a limit value. It records that numerator and denominator both approach zero; their relative behavior still needs analysis.
Replace the expression by an equal formula for nearby inputs. Cancellation is permitted away from the target, which is precisely where the limit is examined.
Cancel a common factor
Evaluate \(\lim_{x\to2}(2x^2-3x-2)/(x^2+x-6)\).
Show worked solution
The simplified denominator tends to 5, so the limit is 5/5=1.
The same form, different outcomes
Compare \(x/x\), \(x^2/x\) and \(x/x^2\) as \(x\to0\).
Hint
Simplify each expression away from zero.
Show worked solution
The first approaches 1; the second approaches 0; the third is 1/x and has opposite unbounded one-sided behavior. The form 0/0 alone determines none of these outcomes.
Squeeze principle
If g(x)≤f(x)≤h(x) for all sufficiently nearby allowed inputs, and g and h approach the same finite L, then f approaches L.
Oscillation with shrinking amplitude
Does \(x\sin(1/x)\) have a limit at zero?
Show worked solution
Since \(|\sin(1/x)|\le1\), the product lies between −|x| and |x|. Both bounds approach zero, so its limit is zero. The oscillating factor alone has no limit; the shrinking amplitude changes the conclusion.
Keep in mind
Bounded does not mean convergent. sin(1/x) remains between −1 and 1 near zero but keeps reaching both extremes.
Rationalizing a radical difference
When a square-root difference creates 0/0, multiplying by its conjugate turns that difference into a difference of squares.
A radical limit
Evaluate \(\lim_{x\to1}(x-1)/(\sqrt{x+3}-2)\).
Show worked solution
The limit is 4. The multiplication uses the conjugate in both numerator and denominator, so it does not change nearby values.
Reverse the quotient
Evaluate \(\lim_{x\to9}(\sqrt{x}-3)/(x-9)\).
Hint
After rationalizing, keep track of which factor remains in the denominator.
Show worked solution
The expression becomes \(1/(\sqrt{x}+3)\) for \(x\ne9\); its limit is \(1/6\).
Infinite limits and vertical asymptotes
Unbounded behavior
\(f(x)\to+\infty\) means outputs eventually exceed every positive bound as the input approaches the target; \(f(x)\to-\infty\) has the corresponding negative meaning. Neither is a finite real limit.
Vertical asymptote
The line \(x=c\) is a vertical asymptote if at least one one-sided limit at \(c\) is \(+\infty\) or \(-\infty\).
The side determines the sign
Find the one-sided limits of \(1/(x-2)\) at 2.
Show worked solution
From the left the denominator is small and negative, so the output tends to −∞. From the right it is small and positive, so the output tends to +∞. The line x=2 is a vertical asymptote.
Keep in mind
A zero denominator is a candidate, not a conclusion. A canceled factor can create a hole; a factor that remains may create an asymptote.
Hole or asymptote?
Classify the exclusions of \(f(x)=(x-1)/((x-1)(x+2))\).
Hint
Keep the original domain after cancellation.
Show worked solution
For x≠1,−2 the rule is 1/(x+2). At x=1 the limit is 1/3, so there is a hole. At x=−2 the one-sided outputs are unbounded, so there is a vertical asymptote.
Limits at infinity and long-run behavior
Limit at infinity
\(\lim_{x\to\infty}f(x)=L\) means outputs approach \(L\) as positive inputs grow without bound. A horizontal asymptote \(y=L\) may arise from either positive or negative infinite input.
For rational functions, divide by the highest denominator power. Lower-order terms vanish; the leading powers control the long-run behavior.
Equal leading degrees
Find \(\lim_{x\to\infty}(5x^3-2x^2+1)/(4x^3+2x-7)\).
Show worked solution
The same ratio holds toward negative infinity, so y=5/4 is the horizontal asymptote.
A sign hidden in a square root
Find \(\lim_{x\to-\infty}\sqrt{9x^6-x}/(x^3+1)\).
Hint
Use √(x⁶)=|x³|, not x³.
Show worked solution
Factor \(|x^3|\) from the numerator. For negative x, \(|x^3|=-x^3\). The ratio therefore approaches −3.
Rationalize an infinite difference
Find \(\lim_{x\to\infty}(\sqrt{x^2+6x}-x)\).
Show worked solution
The limit is 3. Subtracting two quantities that both grow without bound does not determine the limit without examining their difference.
Continuity joins the value to the limit
Continuity at an interior point
\(f\) is continuous at \(c\) when \(f(c)\) is defined, the finite limit \(\lim_{x\to c}f(x)\) exists, and that limit equals \(f(c)\). At a domain endpoint, use the available one-sided limit.
A limit describes approaching a point. Continuity adds the requirement that the actual value belongs to the same behavior. A hole can be repaired by defining one output; a jump cannot.
Repair a hole
Set \(f(x)=(x^2-9)/(x-3)\) for \(x\ne3\) and \(f(3)=k\). Choose k for continuity.
Show worked solution
The nearby limit is 6, so continuity requires k=6. Any other value leaves a discontinuity.
Where is the function continuous?
Find the continuity intervals of \((x^2+4)/(4-25x^2)\).
Hint
A rational function is continuous wherever its denominator is nonzero.
Show worked solution
The exclusions are \(x=\pm2/5\). It is continuous on \((-\infty,-2/5)\), \((-2/5,2/5)\) and \((2/5,\infty)\).
Continuity also lets us make an existence claim without solving an equation. If a continuous temperature record moves from below a target to above it, it must pass through that target; a jump could skip it.
Intermediate Value Theorem
If f is continuous on [a,b] and N lies between f(a) and f(b), then some c in [a,b] satisfies f(c)=N. When N lies strictly between the endpoint values, c lies in (a,b).
A zero exists, but is it unique?
A continuous model has f(0)=−2 and f(3)=5. What can we conclude about f(x)=0?
Show worked solution
At least one solution lies in (0,3). Continuity does not guarantee uniqueness: the curve might cross the axis repeatedly. If it is also strictly increasing, two distinct zeros are impossible, so there is exactly one.
Keep in mind
Checking a few plotted points does not establish continuity. State why the model is continuous before applying the theorem.
Match pieces at their boundaries
A model with several formulas is already continuous inside each piece when those formulas are continuous. The work is at the joining inputs: match the two limits and the assigned value.
Two boundaries, two conditions
Let f(x)=20b+ax−3 for x<−4, f(x)=ax+5 for −4≤x≤−1, and f(x)=bx²−2x+a for x>−1. Choose a and b for continuity.
Show worked solution
At −4: 20b−4a−3=−4a+5, hence b=2/5. At −1: −a+5=b+2+a, hence 2a=3−b=13/5 and a=13/10. Each formula is polynomial, so these two matches establish continuity everywhere.
Keep in mind
Continuity is a property of the whole definition: formula, domain, and assigned boundary values. It cannot be decided from a simplified formula alone.
References
Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen and Christopher J. Stocker, Calculus for Business, Economics, Life Sciences, and Social Sciences, 14th Global Edition, chapters on limits, differentiation and applications. Mark D. Tomenes, Applied Calculus for Business and Economics I, teaching notes and assessment materials (2025–2026).
Further reading: Gilbert Strang and Edwin Herman, Calculus, Volume 1, OpenStax, chapters 2–4.
