Class prep · TJ AP Calculus BC

Learn BC calculus from zero,then practice at test level.

Short lessons that start from zero and build one idea at a time, then original problems at the depth of a hard unit test. Each problem has a full worked solution. Built by Neil Kothari, a TJ alumnus.

Try the free lesson Enroll

$395 covers the full school year. The first lesson is free.

Try one. This is a test-level problem from the course.

Problem 1

Test level
Multiple choice
No calculator
4 pts

Functions $f$ and $g$ satisfy $\displaystyle\lim_{x\to1^-}f(x)=3$ , $\displaystyle\lim_{x\to1^+}f(x)=-1$ , $f(1)=2$ , $\displaystyle\lim_{x\to1^-}g(x)=-1$ , $\displaystyle\lim_{x\to1^+}g(x)=3$ , and $g(1)$ is undefined. Which of the following limits exist? $\text{I. }\lim_{x\to1}\big[f(x)+g(x)\big]\qquad \text{II. }\lim_{x\to1}\big[f(x)\,g(x)\big]\qquad \text{III. }\lim_{x\to1}\frac{f(x)}{g(x)}$

Answer

I and II only. Sum: $3+(-1)=2$ from the left and $-1+3=2$ from the right. Product: $3(-1)=-3$ and $(-1)(3)=-3$ . Quotient: $\frac{3}{-1}=-3$ and $\frac{-1}{3}=-\frac13$ , which differ.

Full solution

Neither $f$ nor $g$ has a two-sided limit at $1$ , so the limit laws (which need each limit to exist) do not apply. Fall back on one-sided limits: each one-sided limit does exist, and one-sided limits obey the sum, product and quotient laws.
The rule that saves you: combine left with left and right with right. Never mix a left-hand value of $f$ with a right-hand value of $g$ .
I. From the left: $3+(-1)=2$ . From the right: $-1+3=2$ . They agree, so $\lim_{x\to1}[f(x)+g(x)]=2$ exists. Graphically, $f$ jumps down by $4$ exactly where $g$ jumps up by $4$ , so the jumps cancel.
II. From the left: $3\cdot(-1)=-3$ . From the right: $(-1)\cdot3=-3$ . They agree, so $\lim_{x\to1}f(x)g(x)=-3$ exists.
III. From the left: $\frac{3}{-1}=-3$ . From the right: $\frac{-1}{3}=-\frac13$ . Neither denominator limit is $0$ , so each one-sided limit is fine, but they differ. The quotient's limit does not exist.
So exactly I and II exist: choice (B). The values $f(1)=2$ and $g(1)$ undefined are distractions; they play no role in any of these limits.
Why the other choices tempt people. (E) comes from believing that if neither part has a limit, no combination can. (A) comes from trusting the sum but assuming a product of two jumps must still jump. (D) comes from pairing the wrong sides (left of $f$ with right of $g$ ): that pairing gives $\frac33=\frac{-1}{-1}=1$ for III and different answers for I and II. (C) comes from extending the pattern without checking III.

Where students lose points

Pairing the wrong sides, or concluding "does not exist" for every combination because $f$ and $g$ each jump. Write a two-column table, left and right, compute each combination in its own column, and compare.

original problems, each with a full worked solution: 780; original problems, each with a full worked solution
at test level or beyond: 540; at test level or beyond
short lessons, one for every idea: 60; short lessons, one for every idea
units, the whole course: 10; units, the whole course

Free lesson

Lesson 1.1: What a Limit Means and How to Read One

Why calculus needs limits, how to read one-sided notation, and how to find limits of single functions, sums, products and compositions straight from a graph or a piecewise formula. Covers College Board AP Calculus BC topics 1.1 to 1.3.

This lesson starts from a simple question and builds up one idea at a time.

Problem 2

Core
True or false
No calculator
2 pts

True or False: A function $g$ can satisfy both $\displaystyle\lim_{x\to 3}g(x)=-1$ and $g(3)=8$ . Justify your answer.

Answer

True. A limit at $3$ is decided only by the values of $g(x)$ for $x$ near $3$ with $x\neq3$ , so the single value $g(3)$ has no say. For example, let $g(x)=2-x$ for $x\neq3$ and $g(3)=8$ . Then $g(x)$ heads to $2-3=-1$ from both sides, so $\lim_{x\to3}g(x)=-1$ , while $g(3)=8$ .

Full solution

Start from what the notation says. $\lim_{x\to3}g(x)=-1$ means you can make $g(x)$ as close to $-1$ as you like by taking $x$ close enough to $3$ , but not equal to $3$ . The phrase in bold is the key to the whole question.
A statement that says something can happen is proved true by one example. So build one.
Begin with a simple function that heads to $-1$ at $x=3$ : the line $y=2-x$ works, since $2-3=-1$ .
Now overrule the single point: define $g(x)=2-x$ for every $x\neq3$ , and set $g(3)=8$ . On a graph, that is the line with an open circle at $(3,-1)$ and a filled dot floating up at $(3,8)$ .
Check the limit from both sides. At $x=2.99$ , $g(x)=-0.99$ ; at $x=3.01$ , $g(x)=-1.01$ . Both sides close in on $-1$ , the left and right limits agree, and so $\lim_{x\to3}g(x)=-1$ . The dot at height $8$ was never used, because $x=3$ itself is excluded.
Check the value: $g(3)=8$ by definition. Both conditions hold at once, so the statement is true.
The big idea: the limit is where the graph is heading; the function value is where the dot is. They match only when the function is continuous at that point, which you will study later in this unit.

Where students lose points

Answering False because "the limit is what you get when you plug in." Plugging in only works for continuous functions. For a limit, cover the point $x=3$ with your finger and read only the nearby values on each side.

Problem 3

Core
Multiple choice
No calculator
4 pts

Which statement means exactly the same thing as $\displaystyle\lim_{x\to-3^{+}}r(x)=6$ ?

Open the rest of this lesson, free

The full lesson and all 13 of its problems, with every solution.

How it works

Class prep that works alongside class

For students currently taking AP Calculus BC. It works alongside class, not in place of it. Students who already know the material can go straight to the test-level problems. Students who feel behind rebuild a topic with the lesson before they try the problems.

Every problem is original, from quick checks to long multi-part problems.
True or false, multiple choice, multi-part and free response, each marked core, test level or beyond, and calculator or no calculator.
Solutions show every step and end with the mistake that most often costs points.
Diagrams wherever a picture makes the idea clear.

What is inside

1Limits and Continuity6 lessons · 78 problems
2Differentiation: Definition and Fundamental Properties6 lessons · 78 problems
3Differentiation: Composite, Implicit, and Inverse Functions6 lessons · 78 problems
4Contextual Applications of Differentiation6 lessons · 78 problems
5Analytical Applications of Differentiation6 lessons · 78 problems
6Integration and Accumulation of Change6 lessons · 78 problems
7Differential Equations6 lessons · 78 problems
8Applications of Integration6 lessons · 78 problems
9Parametric Equations, Polar Coordinates and Vector-Valued Functions6 lessons · 78 problems
10Infinite Sequences and Series6 lessons · 78 problems

The front entrance of Thomas Jefferson High School for Science and Technology

Who built it

Neil Kothari

TJHSST alumnus · Co-founder, EduAvenues · 10+ years in TJ preparation

Neil went to TJ himself and has spent more than a decade working with TJ students and their families. He built this course so students can learn each idea from the beginning and then test themselves on it. Every problem is original and every answer is checked before it goes in.

Tuition

$395for the full school year

One tuition covers every unit of the course. All 10 units are open now. The What a Limit Means and How to Read One lesson stays free for everyone.

10 units open now: 60 lessons and 780 original problems
A full worked solution for every problem
Access through the end of the school year

Enroll