Comment. Continuity quiz
Comment. This quiz is designed to test your knowledge of limits of functions and
Comment. of continuity and uniform continuity.
Shuffle Questions.
Shuffle Answers.
Question. Let X be a subset of R. What does it mean for x to be an adherent point of X?
Correct Answer. For every epsilon > 0, there exists a y in X such that |y-x| < epsilon.
Answer. For every epsilon > 0, there exists a y in X such that 0 < |y-x| < epsilon.
Comment. This is what it means for x to be a limit point of X, not an adherent point.
Answer. There exists epsilon > 0 and y in X such that 0 < |y-x| < epsilon.
Answer. For every epsilon > 0 and all y in X, we have |y-x| < epsilon.
Answer. For every y in X, there exists epsilon > 0 such that |y-x| < epsilon.
Answer. There exists epsilon > 0 such that 0 < |y-x| < epsilon for all y in X.
Answer. There exists y in X such that |y-x| < epsilon for all epsilon > 0.
Question. Let X be a subset of R. What does it mean for x to NOT be an adherent point of X?
Correct Answer. There exists an epsilon > 0 such that |y-x| >= epsilon for all y in X.
Answer. There exists an epsilon > 0 and y in X such that |y-x| >= epsilon.
Answer. For every epsilon > 0 there exists y in X such that |y-x| >= epsilon.
Answer. For every epsilon > 0 we have |y-x| >= epsilon for every y in X.
Answer. For every y in X there exists an epsilon > 0 such that |y-x| >= epsilon.
Answer. There exists y in X such that |y-x| >= epsilon for every epsilon > 0.
Answer. There exists an epsilon > 0 such that y in X whenever |y-x| >= epsilon.
Question. Let X be a subset of R. What does it mean for x to be a limit point of X?
Answer. For every epsilon > 0, there exists a y in X such that |y-x| < epsilon.
Comment. This is what it means for x to be an adherent point of X, not a limit point.
Correct Answer. For every epsilon > 0, there exists a y in X such that 0 < |y-x| < epsilon.
Answer. There exists epsilon > 0 and y in X such that 0 < |y-x| < epsilon.
Answer. For every epsilon > 0 and all y in X, we have |y-x| < epsilon.
Answer. For every y in X, there exists epsilon > 0 such that |y-x| < epsilon.
Answer. There exists epsilon > 0 such that 0 < |y-x| < epsilon for all y in X.
Answer. There exists y in X such that |y-x| < epsilon for all epsilon > 0.
Question. Let X be a subset of R. What does it mean for x to be an adherent point of X?
Correct Answer. There exists a sequence x_1, x_2, x_3, ... in X which converges to x.
Answer. There exists a sequence x_1, x_2, x_3, ... in X-{x} which converges to x.
Comment. This is what it means for x to be a limit point of X, not an adherent point.
Answer. Every sequence x_1, x_2, x_3, ... in X which converges, converges to x.
Answer. Every sequence x_1, x_2, x_3, ... in X-{x} converges to x.
Answer. Every Cauchy sequence x_1, x_2, x_3, ... in X converges to x.
Answer. There exists a sequence x_1, x_2, x_3, ... in X-{x} which converges to x.
Answer. Every sequence x_1, x_2, x_3, ... which converges to x, must lie in X.
Question. Let X be a subset of R. What does it mean for x to be a limit point of X?
Answer. There exists a sequence x_1, x_2, x_3, ... in X which converges to x.
Comment. This is what it means for x to be an adherent point of X, not a limit point.
Correct Answer. There exists a sequence x_1, x_2, x_3, ... in X-{x} which converges to x.
Answer. Every sequence x_1, x_2, x_3, ... in X which converges, converges to x.
Answer. Every sequence x_1, x_2, x_3, ... in X-{x} converges to x.
Answer. Every Cauchy sequence x_1, x_2, x_3, ... in X converges to x.
Answer. Every sequence x_1, x_2, x_3, ... which converges to x, must lie in X.
Answer. Every sequence x_1, x_2, x_3, ... which converges to x, must lie in X - {x}.
Question. Let X be a subset of R, let f: X -> R be a function, and let x be an adherent point of X. What does it mean for lim_{y -> x; y in X} f(x) to equal L?
Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-L| < delta for all y in X for which |y-x| < epsilon.
Answer. For every epsilon > 0, there exists a delta > 0 such that |y-x| < epsilon for all y in X for which |f(y)-L| < delta.
Correct Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-L| < epsilon for all y in X for which |y-x| < delta.
Answer. For every epsilon > 0 and delta > 0, we have |f(y)-L| < epsilon for all y in X for which |y-x| < delta.
Answer. For every epsilon > 0 and every y in X, there exists delta > 0 such that |f(y)-L| < epsilon if |y-x| < delta.
Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-L| < epsilon and |y-x| < delta for some y in X.
Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-L| < epsilon and |y-x| < delta for all y in X.
Question. Let X be a subset of R, let f: X -> R be a function, and let x be an element of X. What does it mean for f to be continuous at x?
Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-f(x)| < delta for all y in X for which |y-x| < epsilon.
Answer. For every epsilon > 0, there exists a delta > 0 such that |y-x| < epsilon for all y in X for which |f(y)-f(x)| < delta.
Correct Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-f(x)| < epsilon for all y in X for which |y-x| < delta.
Answer. For every epsilon > 0 and delta > 0, we have |f(y)-f(x)| < epsilon for all y in X for which |y-x| < delta.
Answer. For every epsilon > 0 and every y in X, there exists delta > 0 such that |f(y)-f(x)| < epsilon if |y-x| < delta.
Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-f(x)| < epsilon and |y-x| < delta for some y in X.
Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-f(x)| < epsilon and |y-x| < delta for all y in X.
Question. Let X be a subset of R, let f: X -> R be a function, and let x be an adherent point of X. What does it mean for lim_{y -> x; y in X} f(x) to equal L?
Correct Answer. For every sequence x_1, x_2, x_3, ... in X which converges to x, the sequence f(x_1), f(x_2), f(x_3), ... converges to L.
Answer. There exists a sequence x_1, x_2, x_3, ... in X converging to x, such that f(x_1), f(x_2), f(x_3), ... converges to L.
Answer. Whenever f(x_1), f(x_2), ... converges to L, the sequence x_1, x_2, x_3, ... , must then converge to x.
Answer. For every sequence x_1, x_2, x_3, ... in X which converges to L, the sequence f(x_1), f(x_2), f(x_3), ... converges to x.
Answer. Whenever f(x_1), f(x_2), ... converges to L, the sequence x_1, x_2, x_3, ... , must lie in X and converge to x.
Answer. Whenever x_1, x_2, ... converges to x, the sequence f(x_1), f(x_2), f(x_3), ... , must lie in X and converge to L.
Answer. For every sequence x_1, x_2, x_3, ... in X which converges to L, the sequence f(x_1), f(x_2), f(x_3), ... converges to x.
Question. Let X be a subset of R, let f: X -> R be a function, and let x be an element of X. What does it mean for f to be continuous at x?
Correct Answer. For every sequence x_1, x_2, x_3, ... in X which converges to x, the sequence f(x_1), f(x_2), f(x_3), ... converges to f(x).
Answer. There exists a sequence x_1, x_2, x_3, ... in X converging to x, such that f(x_1), f(x_2), f(x_3), ... converges to f(x).
Answer. Whenever f(x_1), f(x_2), ... converges to f(x), the sequence x_1, x_2, x_3, ... , must then converge to x.
Answer. Every sequence f(x_1), f(x_2), ... which is convergent, converges to f(x).
Answer. Whenever f(x_1), f(x_2), ... converges to f(x), the sequence x_1, x_2, x_3, ... , must lie in X and converge to x.
Answer. Whenever x_1, x_2, ... converges to x, the sequence f(x_1), f(x_2), f(x_3), ... , must lie in X and converge to f(x).
Answer. For every sequence x_1, x_2, x_3, ... in X, the sequence f(x_1), f(x_2), f(x_3), ... converges to f(x).
Question. Let X be a subset of R, let f: X -> R be a function. What does it mean for f to be continuous on X?
Correct Answer. For every epsilon > 0 and x in X, there exists a delta > 0 such that |f(y)-f(x)| < epsilon for all y in X for which |y-x| < delta.
Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-f(x)| < epsilon for all x, y in X for which |y-x| < delta.
Comment. This is what it means for f to be uniformly continuous on X.
Answer. For every epsilon > 0, there exists a delta > 0 and x,y in X such that |f(y)-f(x)| < delta and |y-x| < epsilon.
Answer. For every epsilon > 0 and x in X, there exists a delta > 0 such that |f(y)-f(x)| < delta for all y in X for which |y-x| < epsilon.
Answer. For every x,y in X, there exists an epsilon > 0 and delta > 0 such that |f(y)-f(x)| < delta and |y-x| < epsilon.
Answer. For every epsilon > 0 and x in X, there exists a delta > 0 such that |y-x| < epsilon for all y in X for which |f(y)-f(x)| < delta.
Answer. For every epsilon > 0, there exists a delta > 0 such that |y-x| < epsilon for all x,y in X for which |f(y)-f(x)| < delta.
Question. Let X be a subset of R, let f: X -> R be a function. What does it mean for f to be uniformly continuous on X?
Answer. For every epsilon > 0 and x in X, there exists a delta > 0 such that |f(y)-f(x)| < epsilon for all y in X for which |y-x| < delta.
Comment. This is what it means for f to be continuous on X.
Correct Answer. For every epsilon > 0, there exists a delta > 0 such that |f(y)-f(x)| < epsilon for all x, y in X for which |y-x| < delta.
Answer. For every epsilon > 0, there exists a delta > 0 and x,y in X such that |f(y)-f(x)| < delta and |y-x| < epsilon.
Answer. For every epsilon > 0 and x in X, there exists a delta > 0 such that |f(y)-f(x)| < delta for all y in X for which |y-x| < epsilon.
Answer. For every x,y in X, there exists an epsilon > 0 and delta > 0 such that |f(y)-f(x)| < delta and |y-x| < epsilon.
Answer. For every epsilon > 0 and x in X, there exists a delta > 0 such that |y-x| < epsilon for all y in X for which |f(y)-f(x)| < delta.
Answer. For every epsilon > 0, there exists a delta > 0 such that |y-x| < epsilon for all x,y in X for which |f(y)-f(x)| < delta.
Question. Let f: [2,4] -> R be a continuous function such that f(2) = 3 and f(4) = 6. The most we can say about the set f([2,4]) is that
Correct Answer. It is a closed interval which contains [3,6].
Answer. It is a set which contains [3,6].
Answer. It is the interval [3,6].
Answer. It is a closed interval.
Answer. It is the interval [2,4].
Answer. It is a set which contains 3 and 6.
Answer. It is a bounded set.
Question. Let f: (1,5) -> R be a continuous function such that f(2) = 3 and f(4) = 6. The most we can say about the set f( (1,5) ) is that
Correct Answer. It is a set which contains [3,6].
Answer. It is a bounded set which contains [3,6].
Answer. It is the interval [3,6].
Answer. It is an open interval.
Answer. It is an interval which contains [3,6].
Answer. It is a set which contains 3 and 6.
Answer. It is a bounded set.
Question. Let f: (1,5) -> R be a uniformly continuous function such that f(2) = 3 and f(4) = 6. The most we can say about the set f( (1,5) ) is that
Answer. It is a set which contains [3,6].
Answer. It is a bounded set which contains [3,6].
Answer. It is the interval [3,6].
Answer. It is an open interval.
Correct Answer. It is an interval which contains [3,6].
Answer. It is an open interval which contains [3,6].
Answer. It is a bounded set.