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Let (xn) be a sequence of real numbers. Suppose the sequence is decreasing and bounded below by 0. What is the most we can deduce from this?
The sequence tends to 0 as n tends to infinity.
The sequence converges as n tends to infinity.
The sequence converges to a non-negative number as n tends to infinity.
We cannot tell whether the sequence converges as n tends to infinity.
Let (xn) be a sequence of real numbers. What does it mean to say that xn tends to a limit x as n tends to infinity?
There is some N so that if n ≥ N then |xn - x| < ε.
If n is big enough, then xn is extremely close to x (in fact, as close as we like).
For any ε > 0, there is some N (which may depend on ε) such that if n ≥ N then |xn - x| < ε.
There is some N so that for any ε > 0, if n ≥ N then |xn - x| < ε.
Let (xn) be a sequence of real numbers. What does it mean to say that xn does not tend to a limit as n tends to infinity?
For any x and any ε > 0, there is N so that if n ≥ N then |xn - x| > ε.
For any x there is some ε > 0 and some N so that if n ≥ N then |xn - x| > ε.
For any x there is some ε > 0 so that for any N if n ≥ N then |xn - x| > ε.
For any x there is some ε > 0 such that for any N there is some n ≥ N with |xn - x| > ε.
Let f be a function from R to R. What does it mean to say that f is continuous at the point a?
For any ε > 0, there is some δ > 0 so that if |x - a| < δ then |f(x) - f(a)| < ε.
For any ε > 0, there is some δ > 0 so that if |x - a| < ε then |f(x) - f(a)| < δ.
There is some δ > 0 so that if |x - a| < δ then |f(x) - f(a)| < ε.
There is some ε > 0 so that for any δ > 0, if |x - a| < δ then |f(x) - f(a)| < ε.
Let xn be a sequence of real numbers. Here are some possible implications about convergence of infinite sums. The sums are all over n from 1 to infinity unless otherwise indicated. Tick any of these implications that are always true.
For each one that's true, can you give a proof? For each one that isn't always true, can you give a counterexample?
Let f be a function from R to R. What does it mean to say that f is not continuous at a?
For any ε > 0, there is δ > 0 so that if |x - a| < δ then |f(x) - f(a)| ≥ ε.
There is some ε > 0 so that for any δ > 0, there is some x with |x - a| < δ and |f(x) - f(a)| ≥ ε.
There is some ε > 0 so that for any δ > 0, if |x - a| < δ then |f(x) - f(a)| ≥ ε.
For any ε > 0, there is δ > 0 and there is some x with |x - a| < δ and |f(x) - f(a)| ≥ ε.
Let (xn) be an infinite sequence of real numbers. Which of the following implications are true? Tick any that are true.
For each statement that's true, can you give a proof? For each that is not always true, can you give a counterexample?
Let f be a continuous function from the interval [a,b] to the real numbers. What is the most we can deduce about f?
f is bounded.
f is bounded and attains its bounds.
f need not be bounded.
f is bounded but need not attain its bounds.
Let f be a function from R to R. What does it mean to say that f(y) tends to l as y tends to x?
There is ε > 0 so that for any δ > 0, if 0 < |y - x| < δ then |f(y) - l| < ε.
There is ε > 0 and there is δ > 0 so that if 0 < |y - x| < δ then |f(y) - l| < ε.
For any ε > 0, there is δ > 0 so that if 0 < |y - x| < δ then |f(y) - l| < ε.
For any ε > 0 and any δ > 0, if 0 < |y - x| < δ then |f(y) - l| < ε.
Let f be a function from [0,1] to R, and suppose that it is twice differentiable on [0,1]. Assume that f(0) =0. What can you deduce about f? Tick any that apply.
For each true statement, give a proof. If the statement need not be true, give a counterexample.