Exercise 5.1 questions and solutions
Exercise 5.1, Question 1
Prove \(f(x) = 5x - 3\) is continuous at \(x = 0\), \(x = -3\) and \(x = 5\).
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- f is a polynomial, so \(\lim_{x \to c} f(x) = 5c - 3 = f(c)\) at every c.
- At 0: limit \(= -3 = f(0)\). At \(-3\): limit \(= -18 = f(-3)\). At 5: limit \(= 22 = f(5)\).
Answer: Continuous at all three points
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Exercise 5.1, Question 2
Examine the continuity of \(f(x) = 2x^2 - 1\) at \(x = 3\).
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- \(\lim_{x \to 3}(2x^2 - 1) = 17 = f(3)\).
Answer: Continuous at \(x = 3\)
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Exercise 5.1, Question 3
Examine for continuity.
(a) \(f(x) = x - 5\)
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- A polynomial: \(\lim_{x \to c} f(x) = c - 5 = f(c)\) for every real c.
Answer: Continuous everywhere
(b) \(f(x) = \dfrac{1}{x - 5}\), \(x \ne 5\)
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- For every \(c \ne 5\): \[\begin{aligned}\lim_{x \to c}\dfrac{1}{x - 5} &= \dfrac{1}{c - 5} \\ &= f(c)\end{aligned}\] The point 5 is not in the domain.
Answer: Continuous at every point of its domain, so a continuous function
(c) \(f(x) = \dfrac{x^2 - 25}{x + 5}\), \(x \ne -5\)
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- For \(x \ne -5\), \(f(x) = x - 5\), and \(\lim_{x \to c} f(x) = c - 5 = f(c)\) for every c in the domain.
Answer: Continuous (on its domain)
(d) \(f(x) = |x - 5|\)
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- For \(c \ne 5\) f is \(\pm(x - 5)\) near c, a polynomial.
- At 5: both one-sided limits are 0 and \(f(5) = 0\).
Answer: Continuous everywhere
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Exercise 5.1, Question 4
Prove \(f(x) = x^n\) is continuous at \(x = n\) (n a positive integer).
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- \(x^n\) is a polynomial: \(\lim_{x \to n} x^n = n^n = f(n)\).
Answer: Proved
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Exercise 5.1, Question 5
Is \(f(x) = \begin{cases}x, & x \le 1 \\ 5, & x > 1\end{cases}\) continuous at \(x = 0\)? At \(x = 1\)? At \(x = 2\)?
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- Near 0 and near 2 f is given by one formula (x, and the constant 5), so it is continuous there.
- At 1: left limit \(= 1\), right limit \(= 5\), \(f(1) = 1\): the limits differ.
Answer: Continuous at 0 and 2; discontinuous at 1
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Exercise 5.1, Question 6
Find all points of discontinuity of f.
\(f(x) = \begin{cases}2x + 3, & x \le 2 \\ 2x - 3, & x > 2\end{cases}\)
Show solution
- Each piece is a polynomial; test only \(x = 2\).
- Left limit \(= 7 = f(2)\), right limit \(= 1\).
Answer: Only \(x = 2\)
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Exercise 5.1, Question 7
\(f(x) = \begin{cases}|x| + 3, & x \le -3 \\ -2x, & -3 < x < 3 \\ 6x + 2, & x \ge 3\end{cases}\)
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- Test \(x = -3\): left limit \(|-3| + 3 = 6\), right limit \(-2(-3) = 6\), \(f(-3) = 6\): continuous.
- Test \(x = 3\): left limit \(-6\), right limit \(20\): discontinuous.
Answer: Only \(x = 3\)
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Exercise 5.1, Question 8
\(f(x) = \begin{cases}\dfrac{|x|}{x}, & x \ne 0 \\ 0, & x = 0\end{cases}\)
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- For \(x < 0\), \(f(x) = -1\); for \(x > 0\), \(f(x) = 1\).
- At 0 the one-sided limits are \(-1\) and \(1\).
Answer: Only \(x = 0\)
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Exercise 5.1, Question 9
\(f(x) = \begin{cases}\dfrac{x}{|x|}, & x < 0 \\ -1, & x \ge 0\end{cases}\)
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- For \(x < 0\), \(\dfrac{x}{|x|} = \dfrac{x}{-x} = -1\). So \(f(x) = -1\) for every x: a constant function.
Answer: No point of discontinuity
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Exercise 5.1, Question 10
\(f(x) = \begin{cases}x + 1, & x \ge 1 \\ x^2 + 1, & x < 1\end{cases}\)
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- At 1: left limit \(1 + 1 = 2\), right limit \(2\), \(f(1) = 2\).
Answer: No point of discontinuity
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Exercise 5.1, Question 11
\(f(x) = \begin{cases}x^3 - 3, & x \le 2 \\ x^2 + 1, & x > 2\end{cases}\)
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- At 2: left limit \(8 - 3 = 5 = f(2)\), right limit \(4 + 1 = 5\).
Answer: No point of discontinuity
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Exercise 5.1, Question 12
\(f(x) = \begin{cases}x^{10} - 1, & x \le 1 \\ x^2, & x > 1\end{cases}\)
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- At 1: left limit \(0 = f(1)\), right limit \(1\).
Answer: Only \(x = 1\)
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Exercise 5.1, Question 13
Is \(f(x) = \begin{cases}x + 5, & x \le 1 \\ x - 5, & x > 1\end{cases}\) continuous?
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- At 1: left limit \(6 = f(1)\), right limit \(-4\).
Answer: No: it is discontinuous at \(x = 1\) (continuous everywhere else)
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Exercise 5.1, Question 14
Discuss the continuity of \(f(x) = \begin{cases}3, & 0 \le x \le 1 \\ 4, & 1 < x < 3 \\ 5, & 3 \le x \le 10\end{cases}\)
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- Constant on each piece; test 1 and 3.
- At 1: left 3, right 4. At 3: left 4, right 5 \(= f(3)\).
Answer: Discontinuous at \(x = 1\) and \(x = 3\); continuous elsewhere on \([0, 10]\)
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Exercise 5.1, Question 15
Discuss the continuity of \(f(x) = \begin{cases}2x, & x < 0 \\ 0, & 0 \le x \le 1 \\ 4x, & x > 1\end{cases}\)
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- At 0: left \(0\), right \(0\), \(f(0) = 0\): continuous.
- At 1: left \(0 = f(1)\), right \(4\): discontinuous.
Answer: Discontinuous only at \(x = 1\)
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Exercise 5.1, Question 16
Discuss the continuity of \(f(x) = \begin{cases}-2, & x \le -1 \\ 2x, & -1 < x \le 1 \\ 2, & x > 1\end{cases}\)
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- At \(-1\): left \(-2 = f(-1)\), right \(-2\). At 1: left \(2 = f(1)\), right \(2\).
Answer: Continuous everywhere
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Exercise 5.1, Question 17
Find the relation between a and b so that \(f(x) = \begin{cases}ax + 1, & x \le 3 \\ bx + 3, & x > 3\end{cases}\) is continuous at \(x = 3\).
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- Left limit and \(f(3)\): \(3a + 1\). Right limit: \(3b + 3\).
- \(3a + 1 = 3b + 3\).
Answer: \(a = b + \tfrac23\) (i.e. \(3a - 3b = 2\))
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Exercise 5.1, Question 18
For what \(\lambda\) is \(f(x) = \begin{cases}\lambda(x^2 - 2x), & x \le 0 \\ 4x + 1, & x > 0\end{cases}\) continuous at \(x = 0\)? What about \(x = 1\)?
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- At 0: left limit \(= \lambda(0) = 0 = f(0)\), right limit \(= 1\). These differ whatever \(\lambda\) is.
- Near \(x = 1\), \(f(x) = 4x + 1\), a polynomial, so f is continuous at 1 for every \(\lambda\).
Answer: No value of \(\lambda\) makes f continuous at 0; f is continuous at \(x = 1\) for every \(\lambda\)
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Exercise 5.1, Question 19
Show \(g(x) = x - [x]\) is discontinuous at every integer.
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- Let n be an integer. For \(n - 1 \le x < n\), \([x] = n - 1\), so the left limit is \(n - (n - 1) = 1\).
- For \(n \le x < n + 1\), \([x] = n\): the right limit and \(g(n)\) are \(n - n = 0\).
- \(1 \ne 0\), so g is discontinuous at n.
Answer: Shown
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Exercise 5.1, Question 20
Is \(f(x) = x^2 - \sin x + 5\) continuous at \(x = \pi\)?
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- \(x^2\), \(\sin x\) and constants are continuous, so their sum is continuous: \[\begin{aligned}\lim_{x \to \pi} f(x) &= \pi^2 - 0 + 5 \\ &= f(\pi)\end{aligned}\]
Answer: Yes
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Exercise 5.1, Question 21
Discuss the continuity of:
(a) \(\sin x + \cos x\)
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- A sum of two functions continuous everywhere.
Answer: Continuous everywhere
(b) \(\sin x - \cos x\)
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- A difference of continuous functions.
Answer: Continuous everywhere
(c) \(\sin x \cdot \cos x\)
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- A product of continuous functions.
Answer: Continuous everywhere
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Exercise 5.1, Question 22
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
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- cos is continuous everywhere: \[\cos(c + h) = \cos c\cos h - \sin c\sin h \to \cos c\] as \(h \to 0\).
- \(\csc x = \dfrac{1}{\sin x}\), \(\sec x = \dfrac{1}{\cos x}\), \(\cot x = \dfrac{\cos x}{\sin x}\): quotients of continuous functions, continuous wherever the denominator is non-zero.
Answer: cos: continuous for all x. cosec and cot: continuous except at \(x = n\pi\); sec: continuous except at \(x = (2n + 1)\tfrac{\pi}{2}\) (\(n \in \mathbb Z\)). Each is continuous on its domain
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Exercise 5.1, Question 23
Find all points of discontinuity of \(f(x) = \begin{cases}\dfrac{\sin x}{x}, & x < 0 \\ x + 1, & x \ge 0\end{cases}\)
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- For \(x < 0\): a quotient of continuous functions with \(x \ne 0\). For \(x > 0\): a polynomial.
- At 0: \(\lim_{x \to 0^-}\dfrac{\sin x}{x} = 1\), right limit \(= 1 = f(0)\).
Answer: None: f is continuous everywhere
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Exercise 5.1, Question 24
Is \(f(x) = \begin{cases}x^2\sin\dfrac1x, & x \ne 0 \\ 0, & x = 0\end{cases}\) continuous?
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- For \(x \ne 0\) it is a product and composite of continuous functions.
- At 0: \[\left|x^2\sin\tfrac1x\right| \le x^2 \to 0\], so the limit is \(0 = f(0)\) (squeeze).
Answer: Yes, continuous everywhere
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Exercise 5.1, Question 25
Examine the continuity of \(f(x) = \begin{cases}\sin x - \cos x, & x \ne 0 \\ -1, & x = 0\end{cases}\)
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- For \(x \ne 0\): continuous (difference of continuous functions).
- At 0: \[\begin{aligned}\lim_{x \to 0}(\sin x - \cos x) &= 0 - 1 \\ &= -1 \\ &= f(0)\end{aligned}\]
Answer: Continuous everywhere
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Exercise 5.1, Question 26
Find k so that f is continuous at the given point.
\(f(x) = \begin{cases}\dfrac{k\cos x}{\pi - 2x}, & x \ne \tfrac{\pi}{2} \\ 3, & x = \tfrac{\pi}{2}\end{cases}\) at \(x = \tfrac{\pi}{2}\)
Show solution
- Put \(x = \tfrac{\pi}{2} + h\): \(\cos x = -\sin h\), \(\pi - 2x = -2h\).
- \[\lim_{h \to 0}\dfrac{-k\sin h}{-2h} = \dfrac{k}{2}\] Set \(\tfrac{k}{2} = 3\).
Answer: \(k = 6\)
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Exercise 5.1, Question 27
\(f(x) = \begin{cases}kx^2, & x \le 2 \\ 3, & x > 2\end{cases}\) at \(x = 2\)
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- \(4k = 3\).
Answer: \(k = \tfrac34\)
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Exercise 5.1, Question 28
\(f(x) = \begin{cases}kx + 1, & x \le \pi \\ \cos x, & x > \pi\end{cases}\) at \(x = \pi\)
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- \(k\pi + 1 = \cos\pi = -1\).
Answer: \(k = -\tfrac{2}{\pi}\)
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Exercise 5.1, Question 29
\(f(x) = \begin{cases}kx + 1, & x \le 5 \\ 3x - 5, & x > 5\end{cases}\) at \(x = 5\)
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- \(5k + 1 = 15 - 5 = 10\).
Answer: \(k = \tfrac95\)
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Exercise 5.1, Question 30
Find a and b so that \(f(x) = \begin{cases}5, & x \le 2 \\ ax + b, & 2 < x < 10 \\ 21, & x \ge 10\end{cases}\) is continuous.
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- At 2: \(2a + b = 5\). At 10: \(10a + b = 21\).
- Subtract: \(8a = 16\).
Answer: \(a = 2,\ b = 1\)
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Exercise 5.1, Question 31
Show \(f(x) = \cos(x^2)\) is continuous.
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- \(f = g \circ h\) with \(h(x) = x^2\) and \(g(x) = \cos x\), both continuous everywhere; a composite of continuous functions is continuous.
Answer: Shown
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Exercise 5.1, Question 32
Show \(f(x) = |\cos x|\) is continuous.
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- \(f = g \circ h\) with \(h(x) = \cos x\) and \(g(x) = |x|\), both continuous.
Answer: Shown
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Exercise 5.1, Question 33
Examine whether \(\sin|x|\) is continuous.
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- \(\sin|x| = g(h(x))\) with \(h(x) = |x|\), \(g(x) = \sin x\): a composite of continuous functions.
Answer: Continuous everywhere
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Exercise 5.1, Question 34
Find all points of discontinuity of \(f(x) = |x| - |x + 1|\).
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- \(|x|\) and \(|x + 1|\) are continuous everywhere, so their difference is too. (The only candidates, \(x = 0\) and \(x = -1\), are corners, not breaks.)
Answer: None
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