Limits and Continuity
Unit 1 of AP Calculus AB, worth 10–12% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Limits from graphs, tables and algebra; limit laws; squeeze theorem; continuity; removable and jump discontinuities; infinite limits and asymptotes; Intermediate Value Theorem.
How this unit is tested
Start by identifying how the limit is presented — graphically, numerically (a table), or algebraically (a formula) — because each format calls for a different first move. On a graph, trace both sides of the input value with your finger; on a table, look at how the outputs trend as the input values close in from below and above; on a formula, try direct substitution first and only dig further if you get an indeterminate form like $0/0$.
When direct substitution fails, factor and cancel, rationalize a radical numerator or denominator, or combine fractions over a common denominator, then substitute again. If none of that works and the expression is squeezed between two functions with the same limit, invoke the squeeze theorem. For limits at infinity or where the denominator approaches zero, reason about the degree of numerator versus denominator and the sign of the expression on each side.
Continuity is a checklist, not a feeling: the limit must exist at the point, the function must be defined there, and the two must be equal. When any one of those three fails, name the failure — a hole (removable), a jump between two finite one-sided limits (jump), or an unbounded blow-up (infinite/essential) — because the AP exam consistently asks you to classify, not just detect.
For the Intermediate Value Theorem, always verify the continuity hypothesis on the closed interval explicitly before concluding a value or root exists; this is a favorite place points are lost for skipping the justification.
What you have to know
Informal definition of a limit
$\lim_{x\to c} f(x) = L$ means the values of $f(x)$ get arbitrarily close to $L$ as $x$ gets close to $c$ from both sides, without requiring $f(c)=L$ or even that $f(c)$ be defined.
Limit laws
If $\lim_{x\to c} f(x)=L$ and $\lim_{x\to c} g(x)=M$, then $\lim_{x\to c}[f(x)\pm g(x)]=L\pm M$, $\lim_{x\to c}[f(x)g(x)]=LM$, $\lim_{x\to c}[kf(x)]=kL$, and $\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{L}{M}$ provided $M\neq 0$.
Squeeze theorem
If $g(x)\le f(x)\le h(x)$ for all $x$ near $c$ (except possibly at $c$), and $\lim_{x\to c} g(x) = \lim_{x\to c} h(x) = L$, then $\lim_{x\to c} f(x) = L$.
Continuity at a point
$f$ is continuous at $x=c$ if all three hold: $f(c)$ is defined, $\lim_{x\to c} f(x)$ exists, and $\lim_{x\to c} f(x) = f(c)$.
Discontinuity classifications
Removable: the limit exists at $c$ but does not equal $f(c)$ (or $f(c)$ is undefined) — appears as a hole. Jump: the left- and right-hand limits at $c$ both exist but are unequal. Infinite (essential): at least one one-sided limit is $\pm\infty$, often at a vertical asymptote.
Intermediate Value Theorem
If $f$ is continuous on the closed interval $[a,b]$ and $N$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c$ in $(a,b)$ such that $f(c)=N$.
14 practice questions
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Evaluate $\lim_{x\to 2} \dfrac{x^2-5x+6}{x-2}$.
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Answer. -1
Direct substitution gives 0/0. Factor the numerator as (x-2)(x-3), cancel the common factor, leaving x-3. Substituting x=2 gives -1. -
A table shows f(0.9)=2.71, f(0.99)=2.97, f(0.999)=2.997, f(1.001)=3.003, f(1.01)=3.03, f(1.1)=3.31. Based on this table, what is $\lim_{x\to 1} f(x)$?
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Answer. 3
As x approaches 1 from both the left and right, f(x) gets arbitrarily close to 3, so the two-sided limit is 3, even though we cannot confirm f(1) itself from a table. -
Use the squeeze theorem to evaluate $\lim_{x\to 0} x^2\sin(1/x)$.
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Answer. 0
Since $-1 \le \sin(1/x) \le 1$, multiplying by $x^2\ge 0$ gives $-x^2 \le x^2\sin(1/x) \le x^2$. Both bounding functions approach 0 as $x\to 0$, so by the squeeze theorem the limit is 0. -
For $f(x) = \begin{cases} x^2+1 & x<2 \\ 3x-1 & x\ge 2 \end{cases}$, is f continuous at x=2?
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Answer. Yes, f is continuous at x=2.
The left-hand limit is $\lim_{x\to2^-}(x^2+1)=5$ and the right-hand limit is $\lim_{x\to2^+}(3x-1)=5$, so the two-sided limit is 5. Since f(2)=3(2)-1=5 matches the limit, all three continuity conditions hold. -
For $f(x) = \begin{cases} x+1 & x<0 \\ x-1 & x\ge 0 \end{cases}$, does $\lim_{x\to 0} f(x)$ exist? What type of discontinuity, if any, occurs at x=0?
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Answer. The limit does not exist; there is a jump discontinuity at x=0.
The left-hand limit is $\lim_{x\to0^-}(x+1)=1$ and the right-hand limit is $\lim_{x\to0^+}(x-1)=-1$. Since these finite one-sided limits disagree, the two-sided limit does not exist, which is the signature of a jump discontinuity. -
Evaluate $\lim_{x\to 3^+} \dfrac{1}{x-3}$ and $\lim_{x\to 3^-} \dfrac{1}{x-3}$, and identify any asymptote.
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Answer. $\lim_{x\to3^+}=+\infty$, $\lim_{x\to3^-}=-\infty$; vertical asymptote at x=3.
As x approaches 3 from the right, x-3 is a small positive number, so 1/(x-3) grows without bound to $+\infty$. From the left, x-3 is a small negative number, so the quotient plunges to $-\infty$. The unbounded behavior on both sides confirms a vertical asymptote at x=3. -
Evaluate $\lim_{x\to\infty} \dfrac{3x^2-5}{2x^2+x-1}$ and state the corresponding horizontal asymptote.
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Answer. 3/2; horizontal asymptote y=3/2.
Numerator and denominator have the same degree (2), so the limit at infinity equals the ratio of leading coefficients, 3/2. This value is also the equation of the horizontal asymptote as x goes to positive or negative infinity. -
If $\lim_{x\to a} f(x) = 4$ and $\lim_{x\to a} g(x) = -2$, find $\lim_{x\to a}[f(x)g(x) + 3g(x)]$.
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Answer. -14
By the product law, $\lim f(x)g(x) = 4\cdot(-2) = -8$. By the constant multiple law, $\lim 3g(x) = 3\cdot(-2) = -6$. Adding these using the sum law gives $-8+(-6) = -14$. -
Let $f(x) = \dfrac{x^2-9}{x-3}$ for $x\ne 3$. What value must f(3) be defined as to make f continuous at x=3?
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Answer. 6
Factor the numerator as (x-3)(x+3), cancel to get x+3 for x≠3, and take the limit as x→3 to get 6. Defining f(3)=6 removes the hole, satisfying all three continuity conditions. -
Show that $f(x) = x^3 - x - 1$ has a root between x=1 and x=2.
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Answer. Yes, by the IVT there is a root in (1,2).
f is a polynomial, hence continuous everywhere including on [1,2]. Since f(1) = 1-1-1 = -1 and f(2) = 8-2-1 = 5, f(1) and f(2) have opposite signs, so 0 lies between them. By the Intermediate Value Theorem, there exists some c in (1,2) with f(c)=0. -
The graph of f has a hole at the point (2,3) and a separate closed dot plotted at (2,5). Find $\lim_{x\to 2} f(x)$ and f(2), and classify the behavior of f at x=2.
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Answer. The limit is 3, f(2)=5, and f has a removable discontinuity at x=2.
The hole in the graph at (2,3) indicates the two-sided limit as x approaches 2 is 3, regardless of what value the function is actually defined to take there. Since f(2) is explicitly plotted as 5, the limit exists but does not equal f(2), which is exactly a removable discontinuity. -
Evaluate $\lim_{x\to 0} \dfrac{\sqrt{x+4}-2}{x}$.
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Answer. 1/4
Direct substitution gives 0/0, so rationalize by multiplying numerator and denominator by the conjugate $\sqrt{x+4}+2$. This turns the numerator into $(x+4)-4 = x$, which cancels with the denominator x, leaving $\dfrac{1}{\sqrt{x+4}+2}$. Substituting x=0 gives $\dfrac{1}{2+2}=\dfrac{1}{4}$. -
Find the value of a that makes $f(x) = \begin{cases} ax+3 & x\le 1 \\ x^2+2 & x>1 \end{cases}$ continuous at x=1.
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Answer. a=0
For continuity, the left-hand value $a(1)+3 = a+3$ must equal the right-hand limit $\lim_{x\to1^+}(x^2+2) = 1+2 = 3$. Setting $a+3=3$ gives $a=0$. -
Explain why $\lim_{x\to 0} \dfrac{1}{x^2}$ does not exist as a finite number, and describe the behavior instead.
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Answer. The limit is $+\infty$; f increases without bound from both sides as x approaches 0.
Since $x^2>0$ for all $x\neq 0$ and approaches 0 as x approaches 0, the fraction $1/x^2$ grows arbitrarily large and positive from both the left and the right. Because both one-sided limits agree and diverge to $+\infty$, we describe this as an infinite limit rather than saying the limit simply fails to exist without explanation.
What people get wrong
- Plugging in the target value and stopping when you get 0/0 instead of recognizing it as a signal to factor, rationalize, or combine fractions before trying again.
- Treating the limit value and the function value f(c) as automatically the same thing — at a removable discontinuity they differ, and the limit exists independently of what f(c) is defined to be.
- Checking only that one-sided limits are equal and forgetting the third continuity condition, that this common value must also equal f(c), which is exactly what makes removable discontinuities easy to miss.
- Getting the sign of an infinite limit wrong by not separately checking the sign of the expression as x approaches from the left versus the right of a vertical asymptote.
- Applying limit laws to a quotient when the denominator's limit is zero, without first checking whether the numerator's limit is also zero (indeterminate, needs more work) or nonzero (limit is infinite or DNE).
- Invoking the Intermediate Value Theorem without first stating that f is continuous on the closed interval in question — the exam awards justification points only when this hypothesis is explicitly verified.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.