Graphs for Calculus:: Limits, Continuity and Discontinuities Flashcards (31 Cards) — Study Online

Curriculum Overview

Learn limits and continuity by reading graphs. 31 flashcards, each with a real graph on the front (drawn with open and filled circles as in a textbook): one-sided and two-sided limits from a graph, holes, jumps and infinite discontinuities, the three-part test for continuity, vertical and horizontal asymptotes, limits at infinity, the Squeeze Theorem, sin(x)/x and the Intermediate Value Theorem. Every answer is checked and explained. For AP Calculus AB/BC, college Calculus 1 and precalculus students.

Topics & Key Concepts

limits continuity discontinuity graphs piecewise one-sided limit asymptote removable jump infinite squeeze theorem intermediate value theorem AP Calculus Calculus 1

Sample Flashcard Questions & Answers

Showing 8 of 31 cards

Question #1 Active Recall

The graph of \(f\) is shown. Find \(\lim_{x \to 1} f(x)\) and \(f(1)\).

Answer & Explanation:

\(\lim_{x \to 1} f(x) = 2\), and \(f(1)\) is **undefined** (the open circle is a hole).

The limit only asks what \(f\) approaches near \(x=1\), not what happens at \(x=1\). Here \(f(x)=\dfrac{x^2-1}{x-1}=x+1\) for \(x \neq 1\).

Question #2 Active Recall

The graph of \(f\) is shown. Find (a) \(\lim_{x \to 2^-} f(x)\), (b) \(\lim_{x \to 2^+} f(x)\), (c) \(\lim_{x \to 2} f(x)\), (d) \(f(2)\).

Answer & Explanation:

(a) \(3\) (b) \(1\) (c) **does not exist** (d) \(f(2)=1\)

Coming from the left the graph heads to the open circle at height 3; from the right it starts at the filled circle at height 1. The one-sided limits differ, so the two-sided limit does not exist. The filled circle gives \(f(2)=1\). Here \(f(x)=x+1\) for \(x

Question #3 Active Recall

The graph of \(f\) is shown. Find \(\lim_{x \to 1} f(x)\) and \(f(1)\). Are they equal?

Answer & Explanation:

\(\lim_{x \to 1} f(x) = 1\) but \(f(1) = 3\) (the filled circle). They are **not equal**.

Both sides of the graph approach the open circle at height 1, so the limit is 1 whatever the single point at \(x=1\) says. The function value at the point is 3. Because the limit and the value differ, \(f\) is not continuous at \(x=1\) (a removable discontinuity). Here \(f(x)=x^2\) for \(x1\), and \(f(1)=3\).

Question #4 Active Recall

The graph of \(f\) is shown. Find \(\lim_{x \to 0^-} f(x)\), \(\lim_{x \to 0^+} f(x)\), \(\lim_{x \to 0} f(x)\) and \(f(0)\).

Answer & Explanation:

\(\lim_{x \to 0^-} f(x) = -1\), \(\lim_{x \to 0^+} f(x) = 1\), \(\lim_{x \to 0} f(x)\) **does not exist**, and \(f(0)\) is **undefined**.

Here \(f(x)=\dfrac{|x|}{x}\), which is \(-1\) for \(x0\). The two open circles at \(x=0\) show that neither height is the value at 0 (the function is not defined there).

Question #5 Active Recall

The graph of \(f(x)=\lfloor x \rfloor\) (the greatest integer function) is shown. Find \(\lim_{x \to 2^-} f(x)\), \(\lim_{x \to 2^+} f(x)\), \(\lim_{x \to 2} f(x)\) and \(f(2)\).

Answer & Explanation:

\(\lim_{x \to 2^-} f(x) = 1\), \(\lim_{x \to 2^+} f(x) = 2\), \(\lim_{x \to 2} f(x)\) **does not exist**, and \(f(2)=2\).

At each integer the graph steps up: the left piece ends in an open circle (height 1 at \(x=2\)) and the right piece begins with a filled circle (height 2). The function takes the value on the right-hand step, \(f(2)=\lfloor 2 \rfloor = 2\).

Question #6 Active Recall

The graph of \(f(x)=\sqrt{x}\) is shown. Find \(\lim_{x \to 0^-} f(x)\), \(\lim_{x \to 0^+} f(x)\) and \(\lim_{x \to 0} f(x)\).

Answer & Explanation:

\(\lim_{x \to 0^-} f(x)\) does **not exist** (there is no graph to the left of 0, \(\sqrt{x}\) is not defined for \(x

Question #7 Active Recall

The graph of \(f(x)=\sin\left(\dfrac{1}{x}\right)\) is shown near \(x=0\). Does \(\lim_{x \to 0} f(x)\) exist?

Answer & Explanation:

**No.** As \(x \to 0\) the function oscillates between \(-1\) and \(1\) faster and faster and never settles on one value, so no limit exists (not even a one-sided limit).

For example \(f(x)=1\) at \(x=\dfrac{1}{\pi/2 + 2\pi n}\) and \(f(x)=-1\) at \(x=\dfrac{1}{3\pi/2 + 2\pi n}\) for every integer \(n \ge 0\), and these points get arbitrarily close to 0.

Question #8 Active Recall

The graph of \(f\) is shown. Find \(\lim_{x \to 0} f(x)\) and say whether \(f(0)\) is defined.

Answer & Explanation:

\(\lim_{x \to 0} f(x) = \dfrac{1}{2}\), and \(f(0)\) is **undefined** (the open circle at height \(0.5\)).

Here \(f(x)=\dfrac{\sqrt{x+1}-1}{x}\). Multiplying top and bottom by \(\sqrt{x+1}+1\) gives \(f(x)=\dfrac{1}{\sqrt{x+1}+1}\) for \(x \neq 0\), which approaches \(\dfrac{1}{1+1}=\dfrac12\).

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