Limits and Continuity Questions and Answers

Answer: A

The true statement is \(f\) is Riemann integrable.

Since \(|x \sin(1/x)| \le |x| \to 0\), \(f\) is continuous at 0, and it is clearly continuous on (0, 1]. A continuous function on [0, 1] is bounded and Riemann integrable.

The other statements fail. \(f\) is continuous at 0, as just shown. It is not differentiable at 0, because \((f(h) - f(0))/h = \sin(1/h)\) has no limit. It is not of bounded variation: between consecutive extrema at \(x = 2/((2k+1)\pi)\) the function swings by about \(4/((2k+1)\pi)\), and these swings form a divergent harmonic-type series.

Answer: D

X contains only one element.

Every subfield of \(\mathbf{R}\) contains 1, hence all integers and all quotients of integers: it contains \(\mathbf{Q}\). So any \(f\) in X vanishes on \(\mathbf{Q}\). Since \(\mathbf{Q}\) is dense in \(\mathbf{R}\) and \(f\) is continuous, \(f(x) = \lim f(q_{n}) = 0\) for every real \(x\), taking rationals \(q_{n} \to x\).

So the only such function is the zero function, and X has exactly one element. The options with more than one element would need a non-zero continuous function vanishing on a dense set, which is impossible.

Answer: A

The answer is that \(f\) is constant on some interval \([a, b]\).

Because \(f\) takes only rational values, its range is countable. For each value \(q\) in the range, the set \(f^{-1}(q)\) is an interval (if \(x \lt y \lt z\) and \(f(x) = f(z) = q\), monotonicity forces \(f(y) = q\)). So R is a union of countably many such intervals. If all of them were single points, R would be countable, which it is not. Hence some \(f^{-1}(q)\) is a non-degenerate interval, and \(f\) is constant on a closed interval \([a, b]\) inside it.

This makes D false, and B false too, since a function constant on an interval is not one-one. C does not hold in general: the floor function \(f(x) = \lfloor x \rfloor\) is increasing, non-constant and rational valued, yet it never takes the value 1/2.

Answer: C

The equation has countably infinitely many solutions.

Let \(g(x) = \cos x - e^{-x}\). For large positive \(x\), \(e^{-x}\) is tiny, so \(g(2k\pi) = 1 - e^{-2k\pi} \gt 0\) and \(g((2k + 1)\pi) = -1 - e^{-(2k+1)\pi} \lt 0\). By the intermediate value theorem there is a root in every interval \((2k\pi, (2k + 1)\pi)\), so there are infinitely many roots.

The roots are countable because \(g\) is a non-constant real-analytic function, whose zeros are isolated, and an isolated set in R is countable. (There is also the root \(x = 0\), and none for \(x \lt 0\), where \(e^{-x} \gt 1\).) So 'no solution', 'finitely many' and 'uncountably many' are all wrong.

Answer: D

Neither \(f\) nor \(g\) is of bounded variation.

\(\sin(1/x)\) oscillates between -1 and 1 infinitely often as \(x \to 0\), so near 0 the variation adds up to 2 on each oscillation and is infinite.

\(x\sin(1/x)\) is continuous but still of unbounded variation near 0. Taking the points \(x_{k} = 1/((k + \frac{1}{2})\pi)\), where \(|g(x_{k})| = x_{k}\), the variation is at least about \(\sum 2x_{k}\), a multiple of the harmonic series, which diverges.

Since both functions have infinite variation on every interval around 0, every option claiming bounded variation for either of them is wrong.

Answer: A

\(g\) is continuous at 0 exactly when u = 2.

Continuity at 0 requires \(u = \lim_{x \to 0} g(x)\). Using \(1 - \cos 2x = 2\sin^{2} x\):

\[ \begin{aligned} \lim_{x \to 0} \frac{1 - \cos 2x}{x^{2}} &= \lim_{x \to 0} \frac{2\sin^{2} x}{x^{2}} \\ &= 2 \left(\lim_{x \to 0} \frac{\sin x}{x}\right)^{2} = 2 \end{aligned} \]

The same value comes from the series \(\cos 2x = 1 - 2x^{2} + \cdots\). So the limit exists and is 2, which rules out 'such a value u does not exist'. Any other value of \(u\), including 1 and 0, leaves a removable discontinuity at 0, because \(g(0)\) would differ from the limit 2. The value 1 is a common slip from \(\cos 2x \approx 1 - x^{2}\), which forgets that the \(x^{2}\) term of \(\cos 2x\) is \((2x)^{2}/2 = 2x^{2}\).

Answer: C

The false statement is if f is differentiable then so is |f|.

Take \(f(x) = x\). It is differentiable everywhere, but \(|f(x)| = |x|\) has a corner at 0: its left derivative there is \(-1\) and its right derivative is \(+1\), so it is not differentiable at 0.

The other statements hold. If \(f\) is continuous, \(|f|\) is the composition of \(f\) with the continuous function \(|\cdot|\). If \(f\) is (Lebesgue) integrable, so is \(|f|\), since integrability is defined through \(\int |f|\). And if \(|f(x)| \le M\) then \(|f|\) is bounded by the same M.

Answer: C

The correct statement is Q = M ∪ N.

Solve \(\frac{1}{4} \lt f(x) \lt \frac{1}{2}\) on each piece.

On \(0 \lt x \le \frac{1}{2}\), \(f(x) = x\), so we need \(\frac{1}{4} \lt x \lt \frac{1}{2}\); the endpoint \(x = \frac{1}{2}\) gives \(f = \frac{1}{2}\), which is not in the open interval. This piece gives M.

On \(\frac{1}{2} \lt x \lt 1\), \(f(x) = x - \frac{1}{2}\), so we need \(\frac{3}{4} \lt x \lt 1\). This piece gives N.

Hence Q is the union of M and N. Q = M misses the second piece, Q contained in N misses the first, and M and N are disjoint, so their intersection is empty.

Answer: D

The uniformly continuous functions are only f₂ and f₃.

\(f_{2}(x) = \cos x\) is continuous on the closed bounded interval \([0, \pi/2]\), and a continuous function on a compact set is uniformly continuous.

\(f_{3}\) is continuous on R (the value 1 at 0 matches \(\lim \sin x / x = 1\)) and tends to 0 as \(x \to \pm\infty\). A continuous function with finite limits at both ends is uniformly continuous on R; its derivative is also bounded, which gives the same conclusion.

\(f_{1}(x) = x^{2}\) on \([0, \infty)\) is not uniformly continuous: with \(x = n\) and \(y = n + 1/n\), \(|x - y| \to 0\) while \(|f_{1}(y) - f_{1}(x)| = 2 + 1/n^{2} \gt 2\). That rules out every option containing \(f_{1}\).

Answer: A

The value is \(\lambda = 1\).

Continuity at \(\pi/2\) needs the limit of \(f\) to equal \(f(\pi/2) = 1\). Put \(h = \pi/2 - x\); then \(\cos x = \cos(\pi/2 - h) = \sin h\), and

\[ \begin{aligned} \lim_{x \to \pi/2} \frac{\lambda \cos x}{\pi/2 - x} &= \lim_{h \to 0} \frac{\lambda \sin h}{h} \\ &= \lambda \end{aligned} \]

Setting \(\lambda = 1\) makes the limit equal the function value. A sign slip (writing \(x - \pi/2\) in the denominator) would give \(-1\), and the values 2 and \(-2\) come from no correct step.