Limits and Continuity Questions and Answers

Answer: B

The necessarily true statement is: \(f\) is bounded and \(f\) has a maximum or a minimum or both.

Because both one-sided limits exist, \(f\) extends to a continuous function on the closed interval \([1,2]\) with the value 3 at both ends. A continuous function on a closed bounded interval is bounded, so \(f\) is bounded and the unbounded option is out.

Now take cases. If \(f \equiv 3\), every point is both a maximum and a minimum. If \(f(x_0) \gt 3\) somewhere, the maximum of the extension exceeds 3, so it is not at an endpoint and is attained inside \((1,2)\). If \(f(x_0) \lt 3\) somewhere, the minimum is attained inside in the same way. So at least one always exists.

'Neither' is never possible. 'One but not both' fails for the constant function, and 'both' is not forced either: \(3 + (x-1)(2-x)\) has a maximum at \(x = \frac{3}{2}\) and no minimum on the open interval.

Answer: D

The necessarily true statement is: there exists \(c \in [1,5]\) such that \(f(c) = f(c+4)\).

Let \(g(x) = f(x+4) - f(x)\) on \([1,5]\). It is continuous, and

\[ \begin{aligned} g(1) &= f(5) - f(1) \\ g(5) &= f(9) - f(5) = -g(1) \end{aligned} \]

since \(f(9) = f(1)\). So \(g(1)\) and \(g(5)\) are zero or of opposite signs, and the intermediate value theorem gives \(c \in [1,5]\) with \(g(c) = 0\). The trick works because 4 is half the length of \([1,9]\).

None of the other three is forced. Take \(f(x) = 1 - x\) on \([1,2]\) and \(f(x) = \frac{x - 9}{7}\) on \([2,9]\): it is continuous, non-constant, and \(f(1) = f(9) = 0\). It is strictly increasing on \([2,9]\), so \(f(c) \lt f(c + h)\) whenever \(2 \leq c\) and \(c + h \leq 9\). No \(c\) in \([4,8]\), \([3,7]\) or \([2,6]\) can work.

Answer: C

\(\sin x\) is Lipschitz and uniformly continuous.

By the mean value theorem, \(\sin x - \sin y = \cos c\,(x - y)\) for some \(c\) between \(x\) and \(y\). Since \(|\cos c| \le 1\),

\[ |\sin x - \sin y| \le |x - y| \]

so sine is Lipschitz with constant 1. Every Lipschitz function is uniformly continuous: given \(\varepsilon \gt 0\), \(\delta = \varepsilon\) works for every pair of points at once.

So the options denying uniform continuity are wrong, and 'uniformly continuous but not Lipschitz' fails because the Lipschitz bound holds. That last description fits a function such as \(\sqrt{x}\) on [0, 1], not \(\sin x\).

Answer: D

The function is discontinuous at all points of R.

This is Dirichlet's function. Rationals and irrationals are both dense in R, so every interval around any point \(a\) contains points where f is 0 and points where f is 1.

If \(a\) is rational, \(f(a) = 1\), but a sequence of irrationals tending to \(a\) has \(f = 0\) all along, so \(f(x_{n}) \not\to f(a)\). If \(a\) is irrational, \(f(a) = 0\), and a sequence of rationals tending to \(a\) has \(f = 1\). Either way f is discontinuous at \(a\).

So it is continuous nowhere: not on R, not at rational points, and not at irrational points. (Continuity at the irrationals only is the property of Thomae's function, which takes the value \(1/q\) at \(p/q\).)

Answer: C

\(f\) is discontinuous at all points except at origin.

At 0: \(|f(x)| \le |x|\) for every \(x\), since \(f(x)\) is either \(x\) or 0. So \(f(x) \to 0 = f(0)\) as \(x \to 0\), and \(f\) is continuous there.

At any \(a \ne 0\): rationals and irrationals are both dense, so there are rational \(x_{n} \to a\) with \(f(x_{n}) = x_{n} \to a\) and irrational \(y_{n} \to a\) with \(f(y_{n}) = 0\). The two limits \(a\) and 0 differ, so \(f\) has no limit at \(a\) and is discontinuous.

That rules out continuity on \(\mathbf{R}\), continuity everywhere except 0 (the reverse of the truth) and discontinuity at every point.

Answer: B

The true statement is there exists \(x \in [0, 1]\) with \(f(x) = x\).

Let \(g(x) = f(x) - x\). Since \(f\) maps into [0, 1], \(g(0) = f(0) \ge 0\) and \(g(1) = f(1) - 1 \le 0\). By the intermediate value theorem \(g\) has a zero, which is a fixed point of \(f\). This is the one-dimensional Brouwer fixed point theorem.

The other statements fail for \(f(x) = 1 - x\). It is not increasing, which refutes A. C fails because its hypothesis holds trivially for \(x = y = 0\) while \(f\) is decreasing. And its only fixed point is 1/2, so having one fixed point does not make \(f\) the identity (D).

Answer: B

The true statement is \(f(x, y) \le |x - y|\) for all \(x, y\).

By the mean value theorem, \(\sin x - \sin y = \cos c\,(x - y)\) for some \(c\) between \(x\) and \(y\), and \(|\cos c| \le 1\). So \(|\sin x - \sin y| \le |x - y|\): sine is Lipschitz with constant 1.

The others fail. (A): \(x = 0, y = \pi/2\) gives \(f = 1 \gt |x| = 0\). (C): \(x = 0, y = \pi\) are different but \(\sin 0 = \sin \pi\), so \(f = 0\). (D): \(x = y = 5\) gives \(f = 0\), which is less than \(|y| = 5\).

Answer: A

The statement that is not true is \(I_{m} f\) is unbounded.

[a, b] is compact and connected, and a continuous image of a compact connected set is compact and connected. In \(\mathbf{R}\) that means a closed bounded interval \([m, M]\), where \(m\) and \(M\) are the minimum and maximum of \(f\) (extreme value theorem).

So the image is an interval (intermediate value theorem), it has a maximum and a minimum, and it is bounded. Only the claim of unboundedness fails.

Answer: C

The correct statement is \(f\) is increasing.

\[ f'(x) = 2x + e^{x} \]

On [0, 1] both terms are non-negative and \(e^{x} \ge 1\), so \(f'(x) \gt 0\) throughout and \(f\) is strictly increasing.

That rules out the others. \(f'\) never vanishes on [0, 1], and a strictly increasing function is one-one. Also \(f(0) = -1 \lt 0\) and \(f(1) = e - 1 \gt 0\), so by the intermediate value theorem \(f(x) = 0\) does have a root in [0, 1].

Answer: B

The answer is \(\min\{\max\{a, b\}, \max\{b, c\}, \max\{a, c\}\}\) (printed without the last comma).

Suppose \(a \lt b \lt c\). Then \(\max\{a, b\} = b\), \(\max\{b, c\} = c\) and \(\max\{a, c\} = c\), and the smallest of these is \(b\), the middle number. The formula is symmetric in \(a, b, c\), so it works for every ordering.

Option A gives \(\min\{b, b, a\} = a\), the smallest. Option C gives \(\max\{b, c, a\} = c\) and option D gives \(c\), the largest. Only B returns the middle value.