AP Calculus AB Unit 1 — Limits and Continuity Worksheet

All 17 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Can Change Occur at an Instant?

   A drone rises from a launch pad. Its height above the pad, in metres, is $h(t)=40t-5t^2$, where $t$ is in seconds.

   1. Find the average rate of change of $h$ over the interval $[1,3]$, with units.
   2. Compute the average rate of change over $[1,1.1]$, over $[1,1.01]$ and over $[1,1.001]$.
   3. Based on part (b), what single number does the drone's velocity at the instant $t=1$ appear to be? Explain in one sentence why no single one of the intervals in part (b) is enough to answer this.
2. Q2Defining Limits and Using Limit Notation

   A function $g$ is defined by $g(x)=2x+5$ for every $x\ne 4$, and $g(4)=1$.

   1. Write, in correct limit notation, the statement “the values of $g(x)$ get arbitrarily close to $13$ as $x$ gets sufficiently close to $4$.”
   2. Evaluate that limit.
   3. Explain in one sentence why the fact that $g(4)=1$ does not change your answer to part (b).
3. Q3Estimating Limit Values from Graphs

   The graph of $f$ consists of three pieces: a line segment rising from $(-2,1)$ to an open circle at $(2,5)$; a solid dot at $(2,2)$; and a horizontal segment from an open circle at $(2,3)$ across to $(6,3)$.

   1. Find ${\displaystyle {lim}_{x\to 2^{-}}f(x)}$ and ${\displaystyle {lim}_{x\to 2^{+}}f(x)}$.
   2. Does ${\displaystyle {lim}_{x\to 2}f(x)}$ exist? Justify your answer.
   3. State $f(2)$.
4. Q4Estimating Limit Values from Tables

   A student tabulates a function $f$ near $x=2$ and records $f(1.9)=0.2516$, $f(1.99)=0.2502$, $f(1.999)=0.2500$, $f(2.001)=0.2500$, $f(2.01)=0.2498$, $f(2.1)=0.2485$. The value $f(2)$ is undefined.

   1. Estimate ${\displaystyle {lim}_{x\to 2}f(x)}$, and say which entries in the list justify treating the limit as two-sided.
   2. Explain in one or two sentences why a table like this can only estimate the limit and can never prove its value.
5. Q5Determining Limits Using Algebraic Properties

   Suppose ${\displaystyle {lim}_{x\to 3}f(x)=-2}$ and ${\displaystyle {lim}_{x\to 3}g(x)=5}$. Find each limit, naming the limit property you use at the decisive step.

   1. ${\displaystyle {lim}_{x\to 3}(3f(x)-2g(x))}$
   2. ${\displaystyle {lim}_{x\to 3}(f(x)g(x)+4)}$
   3. ${\displaystyle {lim}_{x\to 3}\frac{g(x)+3}{f(x)}}$
   4. ${\displaystyle {lim}_{x\to 3}\sqrt{g(x)+11}}$
6. Q6Determining Limits Using Algebraic Manipulation

   Evaluate ${\displaystyle {lim}_{x\to -4}\frac{x^2+x-12}{x+4}}$. Begin by explaining, in one sentence, why substituting $x=-4$ does not settle this limit.

7. Q7Selecting Procedures for Determining Limits

   For each limit, first state which procedure is appropriate — substitution, factoring, or multiplying by a conjugate — and then evaluate it.

   1. ${\displaystyle {lim}_{x\to 2}\frac{x^3-5x}{x+1}}$
   2. ${\displaystyle {lim}_{x\to 5}\frac{x^2-25}{x^2-3x-10}}$
   3. ${\displaystyle {lim}_{x\to 0}\frac{\sqrt{x+16}-4}{x}}$
8. Q8Determining Limits Using the Squeeze Theorem

   Let $f(x)=x^2\cos \left({\displaystyle \frac{1}{x}}\right)$ for $x\ne 0$.

   1. Write an inequality of the form $a(x)\le f(x)\le b(x)$ that holds for every $x\ne 0$, and say what fact about the cosine function justifies it.
   2. Use the squeeze theorem to find ${\displaystyle {lim}_{x\to 0}f(x)}$, stating why the theorem's hypotheses are satisfied.
9. Q9Connecting Multiple Representations of Limits

   Let $f(x)={\displaystyle \frac{x^2-2x-8}{x-4}}$.

   1. Evaluate ${\displaystyle {lim}_{x\to 4}f(x)}$ analytically.
   2. The graph of $f$ is a straight line with a single point missing. Give the equation of that line and the coordinates of the missing point.
   3. Predict the value of $f(3.999)$, to three decimal places, and say how it supports your answer to part (a).
10. Q10Exploring Types of Discontinuities

    Let $f(x)={\displaystyle \frac{(x-1)(x+3)}{(x-1)(x-5)}}$. Identify every value of $x$ at which $f$ is discontinuous, and classify each discontinuity as removable, jump, or infinite. Justify each classification with a limit statement.

11. Q11Defining Continuity at a Point

    Let $f(x)={\displaystyle \frac{x^2-x-2}{x-2}}$ for $x\ne 2$, and $f(2)=3$. Verify, one condition at a time, that $f$ is continuous at $x=2$. Name each of the three conditions as you check it.

12. Q12Confirming Continuity over an Interval

    Find every interval on which $f(x)={\displaystyle \frac{x+1}{x^2-x-6}}$ is continuous. Justify your answer by naming the theorem about where rational functions are continuous.

13. Q13Removing Discontinuities

    Let $$f(x)=\{\begin{array}{cc}{\displaystyle \frac{2x^2+5x+3}{x+1}},\hfill & x\ne -1\hfill \\ [\mathrm{8pt}]k,\hfill & x=-1.\hfill \end{array}$$ Find the value of $k$ that makes $f$ continuous at $x=-1$, and explain why exactly one value works.

14. Q14Connecting Infinite Limits and Vertical Asymptotes

    Let $f(x)={\displaystyle \frac{2x}{x-5}}$.

    1. Find ${\displaystyle {lim}_{x\to 5^{-}}f(x)}$ and ${\displaystyle {lim}_{x\to 5^{+}}f(x)}$, showing the sign reasoning for each.
    2. State the equation of the vertical asymptote, and explain how part (a) establishes it.
    3. Does ${\displaystyle {lim}_{x\to 5}f(x)}$ exist? Justify your answer.
15. Q15Connecting Limits at Infinity and Horizontal Asymptotes
    1. Evaluate ${\displaystyle {lim}_{x\to \infty }\frac{3x^2-5x+1}{6x^2+4}}$ and ${\displaystyle {lim}_{x\to -\infty }\frac{3x^2-5x+1}{6x^2+4}}$.
    2. Evaluate ${\displaystyle {lim}_{x\to \infty }\frac{2x+7}{x^2-1}}$ and ${\displaystyle {lim}_{x\to -\infty }\frac{2x+7}{x^2-1}}$.
    3. State every horizontal asymptote of each of the two functions above.
16. Q16Working with the Intermediate Value Theorem

    Let $f(x)=x^3-4x+1$.

    1. Compute $f(0)$ and $f(1)$.
    2. Show that there is a number $c$ in the open interval $(0,1)$ with $f(c)=0$. Name the theorem you use and verify that each of its hypotheses is satisfied.
17. Q17Synthesis — drawing on several topics in this unit

    Let $f(x)={\displaystyle \frac{x^2-7x+10}{x^2-4x+4}}$.

    1. Write $f$ in lowest terms, stating the values of $x$ excluded from its domain.
    2. Find ${\displaystyle {lim}_{x\to 2^{-}}f(x)}$ and ${\displaystyle {lim}_{x\to 2^{+}}f(x)}$, and give the equation of the vertical asymptote.
    3. Find ${\displaystyle {lim}_{x\to \infty }f(x)}$ and give the equation of the horizontal asymptote.
    4. Classify the discontinuity at $x=2$, and state every interval on which $f$ is continuous.

The 11 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

What a limit claims — and what it never claims

Q2 defines a function by one rule at every input except 4, then deliberately assigns a different value at 4 itself. The limit as x → 4 is unaffected. A limit is a statement about the values of the function at inputs near the point, and the definition excludes the point itself, so whatever is sitting there is not evidence about the limit — not even when it happens to be the right number.

Q1 is where the whole idea comes from. You compute an average rate of change over [1, 3], then over [1, 1.1], [1, 1.01], [1, 1.001]. Every one of those intervals still has positive width, so every answer is still an average — never a velocity at an instant. The instantaneous value is what the averages approach as the width shrinks toward zero, and it is not any one of them. That gap is the reason the course opens here rather than with derivatives.

Q4 makes the point in reverse. Six tabulated values closing in on a number are strong evidence and no proof: a limit is a claim about every input sufficiently close to the point, and a finite list cannot rule out the function doing something else between the entries you happened to sample. On the AP exam, a table supports a conjecture — write "the table suggests", then confirm it algebraically.

A two-sided limit needs both sides to agree

Both one-sided limits must exist and be equal. If they exist and disagree, the two-sided limit does not exist — that is a jump, and no amount of algebra repairs it. This is why Q3 asks for the one-sided limits first: they are the evidence, and the two-sided answer is the conclusion drawn from them.

0/0 is a question, not an answer

Substituting and getting 0/0 tells you the expression must be rewritten before the limit is readable. It tells you nothing about the answer — 0/0 is indeterminate precisely because different functions with that form have different limits. Which rewrite you need is written on the face of the expression.

Q9 asks you to see the same limit three ways: analytically by factoring, graphically as a straight line with one point missing, and numerically at an input very close to the target. They are three views of one fact — and when a hole is what the algebra predicts, the graph has to show a hole exactly there.

Name the property, not just the number

Q5 gives you two limits and asks for four combinations, "naming the limit property you use at the decisive step". The arithmetic is trivial; the mark is on the naming, because two of the four have a hypothesis that must be checked.

Every one of these properties runs in one direction only: if the pieces have limits, then the combination does. Nothing in them licenses the reverse, so a combination with a perfectly good limit tells you nothing about whether its pieces have any.

The squeeze theorem, and when you are forced into it

Q8 asks for the limit of x²cos(1/x) at 0. The product property is unavailable here, and that is the point: cos(1/x) oscillates faster and faster as x approaches 0, so it has no limit there and there is nothing to multiply. What it does have is a bound.

Multiplying an inequality by x² is safe because x² is positive for x ≠ 0 — the direction of the inequalities survives. Then quote the theorem's two hypotheses explicitly: the inequality holds on an open interval around the point (except possibly at the point itself), and the two outer limits both exist and are equal. Only then does the middle function get trapped at that common value. Equal outer limits is the hypothesis students skip, and it is the one doing the work.

Continuity at a point: three conditions, three ways to fail

Q11 asks you to verify continuity "one condition at a time, naming each". Learn them as a numbered list, because the classification of a discontinuity is just a record of which one broke.

Q13 is condition 3 turned into an equation: cancel, take the limit, and set the assigned value equal to it. Exactly one number works, because the limit is a single number — which is also the reason "removable" means removable in one specific way, not adjustable to taste.

Continuity on an interval, and why the answer is usually a union

Q12 wants every interval on which a rational function is continuous, justified by the theorem rather than by inspection. The theorem is short: a rational function is continuous at every point of its domain. So the work is entirely about the domain — factor the denominator, find its zeros, and remove them.

Both kinds of asymptote are limit statements

This is the sentence to memorise, because the two are constantly confused: a vertical asymptote is an infinite limit at a finite input, and a horizontal asymptote is a finite limit at infinity. Different limits, different questions, different method.

For a vertical asymptote (Q14, Q17) the technique is sign analysis, not substitution. Establish that the numerator tends to a nonzero number while the denominator tends to 0, then decide the sign of the denominator just to the left and just to the right of the point. Small negative denominator under a positive numerator gives −∞; small positive gives +∞. The two sides routinely differ, which is exactly why the question asks for both.

For a horizontal asymptote (Q15, Q17), divide the numerator and denominator by the highest power of x in the denominator and let every term with x underneath go to 0. Then check both ends. A horizontal asymptote is a claim about one end of the graph, so x → ∞ and x → −∞ are two questions, and a graph is allowed a different answer at each — or one at only one end.

What the Intermediate Value Theorem actually promises

Q16 asks you to name the theorem and verify each hypothesis. There are exactly two, and both have to be written down:

Read the conclusion carefully, because it is weaker than students expect and stronger than they use. It is an existence claim: it guarantees that such a c exists, never how many there are, and never where. Solving for c afterwards is fine as a check, but the theorem alone gave you only existence — and finding c by algebra is not a use of the IVT. Equally, the theorem never says a function that fails its hypotheses misses the value: drop continuity and the conclusion becomes unavailable, not false.

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Getting the most out of it

Try substitution before you do anything clever

Most limits on this sheet are settled by putting the number in. Substituting first costs one line and tells you which of the four cases you are in: a real number means done, 0/0 means rewrite, and a nonzero number over 0 means you are looking at an infinite limit and should switch to sign analysis.

Write the restriction every time you cancel

"= x + 2 for x ≠ 4" is not pedantry — it is the reason the cancellation is allowed, and it is what keeps the hole in your mental picture of the graph. Students who drop the clause are the ones who later declare a function continuous at a point that is not even in its domain.

Name the theorem out loud

Several questions here award the mark for the justification rather than the value: the limit property at the decisive step, the three conditions of continuity one at a time, the two hypotheses of the IVT. Practise writing the reason as a clause — "the quotient property, and the denominator's limit is nonzero", "continuous on [0, 1] because it is a polynomial". On the free-response section that clause is the point.

Common questions