Limits and Continuity Questions and Answers

Q: Consider the function\[ f(x) = \begin{cases} x \sin\left(\dfrac{1}{x}\right), & \text{if } 0 \lt x \le 1 \\ 0, & \text{if } x = 0 \end{cases} \]Then which of the following statements is true ?

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.

Q: Let \(\{f : \mathbf{R} \to \mathbf{R} : f\) is continuous such that \(f(t) = 0\) for all \(t\) in a subfield of \(\mathbf{R}\} = X\). Then X contains :

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.

Q: Let \(f : \mathbf{R} \to Q\) be a non-constant increasing function. Then :

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.

Q: How many solutions are there for the equation \(\cos x - e^{-x} = 0\) ?

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.

Q: Let \(f, g : \mathbf{R} \to \mathbf{R}\) be defined by \(f(x) = \sin\dfrac{1}{x}\) and \(g(x) = x\sin\dfrac{1}{x}\).Then :

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.

Q: Consider the function \(g\) defined by :\[ g(x) = \begin{cases} \dfrac{1 - \cos(2x)}{x^{2}} & \text{if } x \neq 0 \\ u & \text{if } x = 0 \end{cases} \]The function \(g\) is continuous at 0,

\(g\) 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 \((2x)^{2}/2 = 2x^{2}\).

Q: Let \(f : \mathbf{R} \to \mathbf{R}\) be a function. Then which of the following is not true ?

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.

Q: Suppose that \(f\) is a function defined by :\[ f(x) = \begin{cases} x & \text{if } 0 \lt x \le \tfrac{1}{2} \\ x - \tfrac{1}{2} & \text{if } \tfrac{1}{2} \lt x \lt 1 \end{cases} \]If \(M = \left(\frac{1}{4}, \frac{1}{2}\right)\), \(N = \left(\frac{3}{4}, 1\right)\) and \(Q = \left\{x \mid f(x) \in \left(\frac{1}{4}, \frac{1}{2}\right)\right\}\), then which of the following is true ?

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.

Q: Suppose that \(f_{1}\), \(f_{2}\) and \(f_{3}\) are functions defined by :\[ f_{1}(x) = x^{2},\ 0 \le x \lt \infty \]\[ f_{2}(x) = \cos(x),\ 0 \le x \le \pi/2 \]\[ f_{3}(x) = \begin{cases} \sin(x)/x & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]Which of the above functions is /are uniformly continuous on their respective domains ?

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 \(f_{1}\).

Q: Consider the function f defined by :\[ f(x) = \begin{cases} \dfrac{\lambda \cos(x)}{\pi/2 - x}, & x \neq \pi/2 \\ 1, & x = \pi/2 \end{cases} \]What should be the value of \(\lambda\) so that f would be continuous at \(x = \pi/2\) ?

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.

11.

Consider the function

\[ f(x) = \begin{cases} x \sin\left(\dfrac{1}{x}\right), & \text{if } 0 \lt x \le 1 \\ 0, & \text{if } x = 0 \end{cases} \]

Then which of the following statements is true ?

A.\(f\) is Riemann integrable
B.\(f\) is not continuous at \(x = 0\)
C.\(f\) is of bounded variation
D.\(f\) is differentiable at \(x = 0\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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12.

Let \(\{f : \mathbf{R} \to \mathbf{R} : f\) is continuous such that \(f(t) = 0\) for all \(t\) in a subfield of \(\mathbf{R}\} = X\). Then X contains :

A.uncountable number of elements
B.countably infinite number of points
C.finitely many elements \(\ge 2\)
D.only one element

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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13.

Let \(f : \mathbf{R} \to Q\) be a non-constant increasing function. Then :

A.\(f\) is constant on some interval \([a, b]\)
B.\(f\) is one-one
C.\(f\) is onto
D.\(f\) is not constant on any interval

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

Discuss in Forum

14.

How many solutions are there for the equation \(\cos x - e^{-x} = 0\) ?

A.No solution
B.Finitely many solutions
C.Countably infinite solutions
D.Uncountably many solutions

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

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15.

Let \(f, g : \mathbf{R} \to \mathbf{R}\) be defined by \(f(x) = \sin\dfrac{1}{x}\) and \(g(x) = x\sin\dfrac{1}{x}\).

Then :

A.\(f\) and \(g\) are of bounded variation
B.\(f\) is of bounded variation, but \(g\) is not of bounded variation
C.\(g\) is of bounded variation, but \(f\) is not of bounded variation
D.neither \(f\) nor \(g\) is of bounded variation

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

Discuss in Forum

16.

Consider the function \(g\) defined by :

\[ g(x) = \begin{cases} \dfrac{1 - \cos(2x)}{x^{2}} & \text{if } x \neq 0 \\ u & \text{if } x = 0 \end{cases} \]

The function \(g\) is continuous at 0,

A.if \(u = 2\)
B.if \(u = 1\)
C.if \(u = 0\)
D.such a value \(u\) does not exist

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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 \((2x)^{2}/2 = 2x^{2}\).

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17.

Let \(f : \mathbf{R} \to \mathbf{R}\) be a function. Then which of the following is not true ?

A.If \(f\) is continuous then so is \(|f|\)
B.If \(f\) is integrable then so is \(|f|\)
C.If \(f\) is differentiable then so is \(|f|\)
D.If \(f\) is bounded then so is \(|f|\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

18.

Suppose that \(f\) is a function defined by :

\[ f(x) = \begin{cases} x & \text{if } 0 \lt x \le \tfrac{1}{2} \\ x - \tfrac{1}{2} & \text{if } \tfrac{1}{2} \lt x \lt 1 \end{cases} \]

If \(M = \left(\frac{1}{4}, \frac{1}{2}\right)\), \(N = \left(\frac{3}{4}, 1\right)\) and \(Q = \left\{x \mid f(x) \in \left(\frac{1}{4}, \frac{1}{2}\right)\right\}\), then which of the following is true ?

A.\(Q = M\)
B.\(Q \subseteq N\)
C.\(Q = M \cup N\)
D.\(Q = M \cap N\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

Discuss in Forum

19.

Suppose that \(f_{1}\), \(f_{2}\) and \(f_{3}\) are functions defined by :

\[ f_{1}(x) = x^{2},\ 0 \le x \lt \infty \]

\[ f_{2}(x) = \cos(x),\ 0 \le x \le \pi/2 \]

\[ f_{3}(x) = \begin{cases} \sin(x)/x & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

Which of the above functions is /are uniformly continuous on their respective domains ?

A.None of the three functions
B.Only \(f_{1}\) and \(f_{2}\)
C.Only \(f_{1}\) and \(f_{3}\)
D.Only \(f_{2}\) and \(f_{3}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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 \(f_{1}\).

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20.

Consider the function f defined by :

\[ f(x) = \begin{cases} \dfrac{\lambda \cos(x)}{\pi/2 - x}, & x \neq \pi/2 \\ 1, & x = \pi/2 \end{cases} \]

What should be the value of \(\lambda\) so that f would be continuous at \(x = \pi/2\) ?

A.1
B.−1
C.2
D.−2

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

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.

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