Calculus BC ยท Unit 1
Read a graph the way a multiple-choice limit question is written. Compare one-sided limits, the closed dot, and continuity.
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Read the height the graph approaches
A limit at \(a\) is the y-value the graph approaches. The closed dot is \(f(a)\). An open circle is a height the graph does not use.
One side at a time
\(\lim_{x \to a^-} f(x)\) uses x-values left of \(a\). \(\lim_{x \to a^+} f(x)\) uses x-values right of \(a\).
The two-sided limit exists only when those one-sided limits are equal.
A dashed vertical line
A dashed vertical line is a vertical asymptote. Running off the top means the one-sided limit is \(\infty\). Running off the bottom means \(-\infty\).
A limit can pass inside a continuous function
If \(\lim_{x \to a} f(x) = L\), then \(\lim_{x \to a} \bigl(f(x)\bigr)^2 = L^2\). The value \(f(a)\) is not used.
Read one feature from the graph.
Graph of f
From the left, the graph is the horizontal line at height 1.
The open circle at \((2, 1)\) is the height being approached.
\(\lim_{x \to 2^-} f(x) = 1\)
Answer: \(1\)
Graph of f
From the right, the graph starts at the open circle \((2, -2)\).
\(\lim_{x \to 2^+} f(x) = -2\)
Answer: \(-2\)
Graph of f
The function value is the closed dot.
The closed dot is at \((2, 3)\).
\(f(2) = 3\)
Answer: \(3\)
Graph of f
Both sides approach the open circle at height 6.
The closed dot at height 2 is \(f(3)\), not the limit.
\(\lim_{x \to 3} f(x) = 6\)
Answer: \(6\)
Graph of f
The closed dot is at \((3, 2)\).
\(f(3) = 2\)
Answer: \(2\)
Graph of f
As x approaches 2 from the left, the graph runs off the top of the picture.
The dashed line at \(x = 2\) is a vertical asymptote.
\(\lim_{x \to 2^-} f(x) = \infty\)
Answer: \(\infty\)
Four statements. One of them is true.
Graph of f
\(\lim_{x \to 2^-} f(x) = 1\)
\(\lim_{x \to 2^+} f(x) = -2\)
The one-sided limits are not equal.
\(\lim_{x \to 2} f(x)\) does not exist.
\(f(2) = 3\) is the closed dot, so choices A and C fail.
Answer: D. the two-sided limit does not exist
Graph of f
Both sides approach 6.
\(\lim_{x \to 3} f(x) = 6\)
\(f(3) = 2\), so the limit is not the function value and f is not continuous at 3.
Answer: C. the limit is 6
Graph of f
From the right, the graph runs off the bottom.
\(\lim_{x \to 2^+} f(x) = -\infty\)
From the left, the graph runs off the top, so the two-sided limit is not \(\infty\).
There is no closed dot on the dashed line, so \(f(2)\) is not 0.
Answer: B. the right-hand limit is \(-\infty\)
Graph of f
Both pieces meet at \((1, -1)\), and that point is filled.
\(\lim_{x \to 1^-} f(x) = -1\)
\(\lim_{x \to 1^+} f(x) = -1\)
\(\lim_{x \to 1} f(x) = -1 = f(1)\)
The height 3 is the value at \(x = -3\), not the left-hand limit at 1.
Answer: C. the limit equals the function value
Graph of f
\(\lim_{x \to 3} f(x) = 6\)
Squaring is continuous, so the limit of the square is the square of the limit.
\(\lim_{x \to 3} \bigl(f(x)\bigr)^2 = 6^2\)
\(36\)
\(\bigl(f(3)\bigr)^2 = 2^2 = 4\) is not the limit.
Answer: \(36\)
Graph of f
\(\lim_{x \to 2^-} f(x) = 1\)
\(\lim_{x \to 2^+} f(x) = -2\)
\(f(2) = 3\)
The one-sided limits differ, so the two-sided limit does not exist.
The function value is defined. That does not repair the limit.
Answer: D. f(2) is defined and the two-sided limit does not exist
More than one feature, or a limit of an expression built from f.
Graph of f
\(\lim_{x \to 3^-} f(x) = 6\)
\(\lim_{x \to 3^+} f(x) = 6\)
The two-sided limit exists.
\(f(3) = 2\), which is not 6.
A limit that exists but does not equal the function value is a removable discontinuity.
Answer: B. removable discontinuity at x = 3
Graph of f
\(\lim_{x \to 2^-} f(x) = \infty\)
\(\lim_{x \to 2^+} f(x) = -\infty\)
The one-sided limits are not the same, so the two-sided limit does not exist.
The dashed line is a vertical asymptote.
Answer: A. vertical asymptote, and the two-sided limit does not exist
Graph of f
I. \(\lim_{x \to 1} f(x) = -1\)
II. \(f(1) = -1\)
III. f is continuous at \(x = 1\)
Both sides meet at height \(-1\).
\(\lim_{x \to 1} f(x) = -1\)
The point \((1, -1)\) is filled, so \(f(1) = -1\).
The limit equals the function value, so f is continuous at \(x = 1\).
Answer: D. I, II, and III
Graph of f
\(\lim_{x \to 2^-} f(x) = 1\)
Left-hand limit of \(f(x) - 1\):
\(1 - 1 = 0\)
\(\lim_{x \to 2^+} f(x) = -2\)
Right-hand limit of \(f(x) - 1\):
\(-2 - 1 = -3\)
The one-sided limits are 0 and \(-3\).
The two-sided limit does not exist.
Answer: does not exist
Graph of f
Continuity at 2 needs three facts: \(f(2)\) is defined, the two-sided limit exists, and those two numbers are equal.
\(f(2) = 3\) is defined.
\(\lim_{x \to 2^-} f(x) = 1\) and \(\lim_{x \to 2^+} f(x) = -2\).
The two-sided limit does not exist.
A closed dot is not enough.
Answer: not continuous; the two-sided limit does not exist
Graph of f
\(\lim_{x \to 3} f(x) = 6\)
Multiply by 2, then subtract 5. Each step uses the limit, not \(f(3)\).
\(2 \cdot 6 = 12\)
\(12 - 5 = 7\)
Answer: \(7\)
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