Solution. In each case the method is the same: express \(\left |f(x)-L\right |\) in terms of \(\left |x-c\right |\) and make the result smaller than \(\varepsilon \).

(a)

\[\left |2x-3-1\right | = \left |2x-4\right | = 2\left |x-2\right | < 0.001 \implies \left |x-2\right | < 0.0005 .\] So \(\delta = 0.0005\) works.

(b)

For \(x\neq -3\), \(\dfrac {x^2-9}{x+3} = x-3\), so \[\left |(x-3)-(-6)\right | = \left |x+3\right | < 0.005 ,\] and \(\delta = 0.005\) works. The removable discontinuity at \(x=-3\) is irrelevant, since the definition never uses the value at \(x=c\).

(c)

\[\left |\frac {1}{x} - \frac {1}{2}\right | = \frac {\left |2-x\right |}{2\left |x\right |}.\] The factor \(\dfrac {1}{\left |x\right |}\) must be controlled first. Restricting \(\left |x-2\right |<1\) gives \(1<x<3\), hence \(2\left |x\right |>2\) and \[\left |\frac {1}{x}-\frac {1}{2}\right | < \frac {\left |x-2\right |}{2} .\] Requiring this to be below \(0.002\) gives \(\left |x-2\right |<0.004\), so \(\delta = \min \{1, 0.004\} = 0.004\).

(d)

\[\left |\frac {4}{x+3} - 1\right | = \frac {\left |4 - (x+3)\right |}{\left |x+3\right |} = \frac {\left |1-x\right |}{\left |x+3\right |}.\] Restricting \(\left |x-1\right |<1\) gives \(0<x<2\), so \(\left |x+3\right |>3\) and the quotient is less than \(\dfrac {\left |x-1\right |}{3}\). Requiring this below \(0.001\) gives \(\left |x-1\right |<0.003\), so \(\delta = \min \{1,0.003\} = 0.003\).

Solution.

(a)

\(\left |3x-5+2\right | = 3\left |x-1\right | < \varepsilon \) exactly when \(\left |x-1\right | < \dfrac {\varepsilon }{3}\), so take \(\delta = \dfrac {\varepsilon }{3}\).

(b)

For \(x\neq 2\), \(\dfrac {x^2-4}{x-2} = x+2\), so \(\left |(x+2)-4\right | = \left |x-2\right | < \varepsilon \) and \(\delta = \varepsilon \) works.

(c)

\[\left |\frac {1}{x+3} - 1\right | = \frac {\left |1-(x+3)\right |}{\left |x+3\right |} = \frac {\left |x+2\right |}{\left |x+3\right |}.\] Restricting \(\left |x+2\right | < \dfrac 12\) gives \(\left |x+3\right | > \dfrac 12\), so the quotient is below \(2\left |x+2\right |\). Take \(\delta = \min \Big \{\dfrac 12, \dfrac {\varepsilon }{2}\Big \}\).

(d)

\[\left |\frac {x-1}{x+1} + 1\right | = \left |\frac {x-1+x+1}{x+1}\right | = \frac {2\left |x\right |}{\left |x+1\right |}.\] Restricting \(\left |x\right | < \dfrac 12\) gives \(\left |x+1\right | > \dfrac 12\), so the quotient is below \(4\left |x\right |\). Take \(\delta = \min \Big \{\dfrac 12, \dfrac {\varepsilon }{4}\Big \}\).

Solution.

(a)

Rationalise the numerator: \[\frac {\sqrt {x+h}-\sqrt {x}}{h}\cdot \frac {\sqrt {x+h}+\sqrt {x}}{\sqrt {x+h}+\sqrt {x}} = \frac {h}{h\big (\sqrt {x+h}+\sqrt {x}\big )} = \frac {1}{\sqrt {x+h}+\sqrt {x}}\longrightarrow \frac {1}{2\sqrt {x}} .\]

(b)

Factor: \(x^3-8 = (x-2)(x^2+2x+4)\), so the quotient is \(x^2+2x+4\rightarrow 12\).

(c)

Both parts vanish at \(x=2\). Rationalise the denominator: \[\frac {4-x^2}{3-\sqrt {x^2+5}}\cdot \frac {3+\sqrt {x^2+5}}{3+\sqrt {x^2+5}} = \frac {\big (4-x^2\big )\big (3+\sqrt {x^2+5}\big )}{9-\big (x^2+5\big )} = \frac {\big (4-x^2\big )\big (3+\sqrt {x^2+5}\big )}{4-x^2}.\] Cancelling gives \(3+\sqrt {x^2+5}\rightarrow 3+3 = 6\).

(d)

As \(x\rightarrow 1^+\) the numerator tends to \(1\) while \(1-x\rightarrow 0\) through negative values, so the quotient tends to \(-\infty \).

(e)

The function is continuous at \(x=2\), so substitute: \(\sqrt {\dfrac {9}{4}} = \dfrac {3}{2}\).

(f)

Divide numerator and denominator by \(x^3\): \[\frac {\frac {2}{x} + \frac {7}{x^2} + \frac {5}{x^3}}{1 + \frac {2}{x^2} + \frac {1}{x^3}} \longrightarrow \frac {0}{1} = 0 .\] The denominator has the higher degree, so the limit is \(0\).

(g)

Here the numerator has the higher degree. Dividing by \(x\), \[\frac {x + \frac {4}{x}}{1 + \frac {2}{x}} \longrightarrow -\infty \quad \text {as } x\rightarrow -\infty .\]

(h)

As \(x\rightarrow 0^-\) the numerator tends to \(1\) and \(x\rightarrow 0\) through negative values, so the quotient tends to \(-\infty \).

Solution. This is the derivative computed from first principles.

(a)

\[\frac {\big [(x+h)^2 - 3(x+h)\big ] - \big [x^2-3x\big ]}{h} = \frac {2xh + h^2 - 3h}{h} = 2x + h - 3 \longrightarrow 2x-3 .\]

(b)

Rationalising the numerator, \[\frac {\sqrt {5x+5h+1}-\sqrt {5x+1}}{h} = \frac {5h}{h\big (\sqrt {5x+5h+1}+\sqrt {5x+1}\big )} \longrightarrow \frac {5}{2\sqrt {5x+1}} .\]

Solution.

(a)

Not indeterminate: substituting gives \(\dfrac {4+2-6}{4-6} = \dfrac {0}{-2} = 0\). A zero numerator with a non-zero denominator is simply zero.

(b)

Of the form \(\dfrac {0}{0}\). Dividing above and below by \(x\), \[\frac {1 - \frac {\sin 2x}{x}}{1 + \frac {\sin 2x}{x}} \longrightarrow \frac {1-2}{1+2} = -\frac {1}{3},\] using \(\dfrac {\sin 2x}{x}\rightarrow 2\).

(c)

Of the form \(0\cdot \infty \). Write it as a quotient: \[\frac {1-\tan x}{\cos 2x} \qquad \Big (\tfrac 00\Big ),\] and apply l’Hôpital: \[\frac {-\sec ^2x}{-2\sin 2x} = \frac {\sec ^2 x}{2\sin 2x} \longrightarrow \frac {2}{2} = 1 .\]

(d)

Of the form \(1^{\infty }\). Taking logarithms, with \(y = x^{1/(x-1)}\), \[\ln y = \frac {\ln x}{x-1} \qquad \Big (\tfrac 00\Big ) \longrightarrow \frac {1/x}{1} = 1 ,\] so \(y\rightarrow e\).

(e)

Of the form \(0\cdot (-\infty )\). Write \(x^2\ln x = \dfrac {\ln x}{x^{-2}}\), of type \(\dfrac {-\infty }{\infty }\), and apply l’Hôpital: \[\frac {1/x}{-2x^{-3}} = -\frac {x^2}{2}\longrightarrow 0 .\]

(f)

Of the form \(\infty -\infty \). Combine over a common denominator: \[\csc x - \cot x = \frac {1-\cos x}{\sin x} \qquad \Big (\tfrac 00\Big ) \longrightarrow \frac {\sin x}{\cos x} = \tan x \longrightarrow 0 .\]

(g)

Of the form \(\infty ^0\). With \(y = (\tan x)^{\cos x}\), \[\ln y = \cos x\ln \tan x = \frac {\ln \tan x}{\sec x} \qquad \Big (\tfrac {\infty }{\infty }\Big ).\] Differentiating, \[\frac {\sec ^2x/\tan x}{\sec x\tan x} = \frac {\sec x}{\tan ^2 x} = \frac {\cos x}{\sin ^2 x}\longrightarrow 0 ,\] so \(y\rightarrow e^0 = 1\).

Solution.

(a)

\(f(x) = 1\) for \(x>0\) and \(-1\) for \(x<0\), undefined at \(x=0\). The one-sided limits are \(1\) and \(-1\), so no limit exists at \(0\): a jump discontinuity, and not removable.

(b)

\(\dfrac {x^2-3x-10}{x+2} = \dfrac {(x-5)(x+2)}{x+2} = x-5\) for \(x\neq -2\). The limit at \(-2\) is \(-7\) but the function is undefined there: a removable discontinuity, repaired by setting \(f(-2) = -7\).

(c)

\(\dfrac {x^4-1}{x^2-1} = \dfrac {(x^2-1)(x^2+1)}{x^2-1} = x^2+1\) for \(x\neq \pm 1\). Both \(x=1\) and \(x=-1\) are removable, repaired by setting \(f(\pm 1) = 2\).

(d)

The pieces meet at \(x=2\). From the left \(x^2+1\rightarrow 5\); from the right \(x+3\rightarrow 5\); and \(f(2) = 5\). The limits agree with the value, so \(f\) is in fact continuous at \(2\) and has no discontinuity at all.

(e)

Two junctions. At \(x=0\): from the left \(x-1\rightarrow -1\), which is \(f(0)\); from the right \(x-2\rightarrow -2\). The one-sided limits differ, so there is a jump, not removable. At \(x=3\): from the left \(x-2\rightarrow 1\); from the right \(4-x\rightarrow 1\), and \(f(3) = 1\). Continuous there.

Solution. Rolle’s theorem requires \(f\) continuous on \([a,b]\), differentiable on \((a,b)\) and \(f(a) = f(b)\); it then supplies \(x_0\in (a,b)\) with \(f'(x_0) = 0\).

(a)

\(f(0) = 0\) and \(f(2\sqrt 3) = 24\sqrt 3 - 24\sqrt 3 = 0\), so the hypotheses hold. \[f'(x) = 3x^2-12 = 0 \implies x = \pm 2 ,\] and only \(x_0 = 2\) lies in \(\big (0, 2\sqrt 3\big )\).

(b)

\(\sin 0 = \sin \pi = 0\) and \(f'(x) = \cos x = 0\) gives \(x_0 = \dfrac {\pi }{2}\), which lies in \((0,\pi )\).

(c)

\(\cos \dfrac {\pi }{2} = \cos \dfrac {3\pi }{2} = 0\) and \(f'(x) = -\sin x = 0\) gives \(x = 0\) or \(\pi \); only \(x_0 = \pi \) lies in the interval.

Solution. Both vanish at \(x=0\) and \(x=4\), so \([0,4]\) is the natural interval to test.

(a)

No. The function is undefined at \(x=2\), which lies inside \((0,4)\), so it is neither continuous on \([0,4]\) nor differentiable on \((0,4)\). The hypotheses fail and the conclusion need not hold.

(b)

Yes. The only singularity is at \(x=-2\), outside \([0,4]\), so \(f\) is continuous and differentiable throughout and \(f(0) = f(4) = 0\). Differentiating, \[f'(x) = \frac {(2x-4)(x+2) - \big (x^2-4x\big )}{(x+2)^2} = \frac {x^2+4x-8}{(x+2)^2},\] which vanishes when \(x^2+4x-8 = 0\), that is \(x = -2\pm 2\sqrt 3\). The root \(x_0 = -2+2\sqrt 3 \approx 1.46\) lies in \((0,4)\), as Rolle guarantees.

Solution. The theorem gives \(x_0\in (a,b)\) with \(f'(x_0) = \dfrac {f(b)-f(a)}{b-a}\).

(a)

\(f(3) = 36\), \(f(1) = 4\), so the slope is \(\dfrac {36-4}{2} = 16\). Then \(f'(x) = 6x+4 = 16\) gives \(x_0 = 2\).

(b)

The slope is \(\dfrac {216-0}{6} = 36\), and \(f'(x) = 3x^2 = 36\) gives \(x = \pm 2\sqrt 3\); only \(x_0 = 2\sqrt 3 \approx 3.46\) lies in \((0,6)\).

(c)

The slope is \(\dfrac {\ln 2e - \ln 1}{2e-1} = \dfrac {1+\ln 2}{2e-1}\), and \(f'(x) = \dfrac 1x\) gives \[x_0 = \frac {2e-1}{1+\ln 2} \approx 2.62 ,\] which lies in \((1, 2e)\).

Solution. The theorem gives \(f(b) - f(a) = f'(c)(b-a)\) for some \(c\) between \(a\) and \(b\). Choosing \(a\) where \(f\) is known exactly and \(b\) the target, and approximating \(f'(c)\) by its value at the known endpoint, gives the estimates below.

(a)

Take \(f(x) = \sqrt x\) on \([15,16]\). Then \(4 - \sqrt {15} = \dfrac {1}{2\sqrt c}\) with \(15<c<16\), so \[\sqrt {15} \approx 4 - \frac {1}{2\sqrt {16}} = 4 - \frac 18 = 3.875 ,\] against the true value \(3.8730\).

(b)

Take \(f(x) = x^{1/6}\) on \([64,65]\), so \(f'(x) = \dfrac {1}{6}x^{-5/6}\) and \[\sqrt [6]{65} \approx 2 + \frac {1}{6\cdot 32} = 2 + \frac {1}{192} \approx 2.00521 ,\] against \(2.00517\).

(c)

Take \(f(x) = x^3\) on \([3, 3.001]\), so \(f'(x) = 3x^2\) and \[(3.001)^3 \approx 27 + 3(9)(0.001) = 27.027 ,\] against \(27.027009\).

(d)

Take \(f(x) = \dfrac 1x\) on \([999,1000]\), so \(f'(x) = -\dfrac {1}{x^2}\) and \[\frac {1}{999} \approx \frac {1}{1000} + \frac {1}{1000^2} = 0.001001 ,\] against \(0.001001001\ldots \)

Solution. Fix \(x\) with \(0<x<\dfrac {\pi }{2}\) and apply the theorem to \(f(t) = \tan t\) on \([0,x]\), where it is continuous and differentiable. There is \(c\in (0,x)\) with \[\frac {\tan x - \tan 0}{x - 0} = \sec ^2 c , \qquad \text {that is}\qquad \tan x = x\sec ^2 c .\] Since \(0<c<\dfrac {\pi }{2}\) we have \(0<\cos c<1\), hence \(\sec ^2 c>1\), and as \(x>0\), \[\tan x = x\sec ^2c > x .\]

Solution. Cauchy’s form states that \[\frac {f(b)-f(a)}{g(b)-g(a)} = \frac {f'(x_0)}{g'(x_0)}\] for some \(x_0\in (a,b)\), provided \(g'\) does not vanish there. Here \[\frac {f(4)-f(1)}{g(4)-g(1)} = \frac {14-5}{17-2} = \frac {9}{15} = \frac {3}{5},\] while \(\dfrac {f'(x)}{g'(x)} = \dfrac {3}{2x}\). Setting these equal, \[\frac {3}{2x_0} = \frac {3}{5} \implies x_0 = \frac {5}{2},\] which lies in \((1,4)\). Note \(g'(x) = 2x\neq 0\) throughout, as required.

Solution. A geometric series \(\sum ar^{n-1}\) converges precisely when \(\left |r\right |<1\), to \(\dfrac {a}{1-r}\).

(a)

\(a = 1\), \(r = \dfrac 12\). Since \(\left |r\right |<1\) it converges to \(\dfrac {1}{1-\frac 12} = 2\).

(b)

\(a = 4\), \(r = -\dfrac 14\). Since \(\left |r\right |<1\) it converges to \(\dfrac {4}{1+\frac 14} = \dfrac {4}{\frac 54} = \dfrac {16}{5}\).

(c)

\(a = 1\), \(r = \dfrac 32 > 1\), so the series diverges.

Solution.

(a)

Geometric with \(a = \dfrac 13\), \(r = \dfrac 13\), so the sum is \(\dfrac {1/3}{1-1/3} = \dfrac 12\).

(b)

Partial fractions give \(\dfrac {1}{(2n-1)(2n+1)} = \dfrac 12\Big (\dfrac {1}{2n-1} - \dfrac {1}{2n+1}\Big )\), so the partial sum telescopes: \[S_N = \frac 12\Big (1 - \frac {1}{2N+1}\Big )\longrightarrow \frac 12 .\]

(c)

Here \(\dfrac {1}{n(n+2)} = \dfrac 12\Big (\dfrac 1n - \dfrac {1}{n+2}\Big )\), and the terms cancel in pairs two apart, leaving the first two: \[S_N = \frac 12\Big (1 + \frac 12 - \frac {1}{N+1} - \frac {1}{N+2}\Big ) \longrightarrow \frac 12\cdot \frac 32 = \frac 34 .\]

(d)

Here \[\frac {1}{n(n+1)(n+2)} = \frac {1}{2n} - \frac {1}{n+1} + \frac {1}{2(n+2)} ,\] and telescoping leaves \[S_N \longrightarrow \frac 12 - \frac 12 + \frac 14 = \frac 14 .\]

Solution. In each case the associated function is positive, continuous and eventually decreasing on \([1,\infty )\), so the test applies and the series converges exactly when the integral does.

(a)

\(\displaystyle {\int _1^{\infty }\frac {dx}{x} = \lim _{b\rightarrow \infty }\ln b = \infty }\), so the harmonic series diverges.

(b)

\(\dfrac {50}{x(x+1)} = 50\Big (\dfrac 1x - \dfrac {1}{x+1}\Big )\), so \[\int _1^{\infty }\frac {50\,dx}{x(x+1)} = 50\Big [\ln \frac {x}{x+1}\Big ]_1^{\infty } = 50\big (0 - \ln \tfrac 12\big ) = 50\ln 2 ,\] finite, so the series converges.

(c)

\[\int _1^{\infty }\frac {x\,dx}{x^2+1} = \Big [\tfrac 12\ln \big (x^2+1\big )\Big ]_1^{\infty } = \infty ,\] so the series diverges. Note the terms tend to \(0\), which by itself proves nothing.

(d)

\[\int _1^{\infty }\frac {dx}{(2x+1)^2} = \Big [-\frac {1}{2(2x+1)}\Big ]_1^{\infty } = \frac 16 ,\] finite, so the series converges.

Solution.

(a)

For \(n\geq 2\), \(n^3 - 1 \geq \dfrac {n^3}{2}\), so \(\dfrac {1}{n^3-1}\leq \dfrac {2}{n^3}\). Since \(\sum \dfrac {1}{n^3}\) is a \(p-\)series with \(p = 3 > 1\) it converges, so the given series converges.

(b)

\(\dfrac {1}{\sqrt [3]{n}} = \dfrac {1}{n^{1/3}}\) is a \(p-\)series with \(p = \dfrac 13 \leq 1\), so it diverges.

(c)

For \(n\geq 3\), \(n^{n-1}\geq 2^{n-1}\), so \(\dfrac {1}{n^{n-1}}\leq \dfrac {1}{2^{n-1}}\), the terms of a convergent geometric series. Hence it converges.

(d)

For large \(n\), \(\dfrac {n}{3n^2-4}\) behaves like \(\dfrac {1}{3n}\). Precisely, for \(n\geq 2\), \(3n^2 - 4 \leq 3n^2\), so \[\frac {n}{3n^2-4} \geq \frac {n}{3n^2} = \frac {1}{3n},\] and \(\sum \dfrac {1}{3n}\) diverges. Hence the series diverges.

Solution. The test examines \(L = \lim \left |\dfrac {a_{n+1}}{a_n}\right |\): convergence if \(L<1\), divergence if \(L>1\), no conclusion if \(L=1\).

(a)

\[\frac {a_{n+1}}{a_n} = \frac {(n+2)(n+3)}{(n+1)!}\cdot \frac {n!}{(n+1)(n+2)} = \frac {n+3}{(n+1)^2}\longrightarrow 0 < 1 ,\] so it converges.

(b)

\[\frac {a_{n+1}}{a_n} = \frac {5^{n+1}}{(n+1)!}\cdot \frac {n!}{5^n} = \frac {5}{n+1}\longrightarrow 0 < 1 ,\] so it converges.

(c)

\[\frac {a_{n+1}}{a_n} = \frac {n+1}{2^{2n+2}}\cdot \frac {2^{2n}}{n} = \frac {1}{4}\cdot \frac {n+1}{n}\longrightarrow \frac 14 < 1 ,\] so it converges.

(d)

\[\frac {a_{n+1}}{a_n} = \frac {(n+1)^{n+1}}{(n+1)!}\cdot \frac {n!}{n^n} = \frac {(n+1)^n}{n^n} = \Big (1+\frac 1n\Big )^n\longrightarrow e > 1 ,\] so it diverges.

Solution. The alternating series test requires the absolute values to decrease monotonically to \(0\); both conditions are needed.

(a)

\(a_n = \dfrac {1}{n^2}\) decreases and tends to \(0\), so it converges. (It also converges absolutely, being a \(p-\)series with \(p=2\).)

(b)

The denominators are \(n^2+1\), so \(a_n = \dfrac {1}{n^2+1}\), which decreases to \(0\): it converges.

(c)

\(a_n = \dfrac {1}{n!}\) decreases to \(0\), so it converges. Its sum is \(1 - e^{-1}\).

(d)

\(a_n = \dfrac {1}{2n-1}\) decreases to \(0\), so it converges. This is Leibniz’s series, with sum \(\dfrac {\pi }{4}\).

Solution.

(a)

The absolute series \(\sum \dfrac {1}{(2n-1)^3}\) converges by comparison with \(\sum \dfrac {1}{n^3}\), so the series is absolutely convergent.

(b)

\(\dfrac {n}{n+1}\rightarrow 1 \neq 0\), so the terms do not tend to zero and the series diverges outright. Neither form of convergence applies, and the alternating series test never gets started.

(c)

The absolute series \(\sum \dfrac {1}{(n+1)^2}\) converges, so this is absolutely convergent.

(d)

The absolute series \(\sum \dfrac {1}{3n-1}\) behaves like \(\sum \dfrac {1}{3n}\) and diverges. But \(\dfrac {1}{3n-1}\) decreases to \(0\), so the alternating series itself converges. Hence it is conditionally convergent.

Solution.

(a)

Here \(a_n = nx^n\), so \[\left |\frac {a_{n+1}}{a_n}\right | = \frac {n+1}{n}\left |x\right | \longrightarrow \left |x\right | ,\] giving convergence for \(\left |x\right |<1\). At \(x = \pm 1\) the terms are \(\pm n\), which do not tend to \(0\), so both endpoints fail. The interval is \((-1, 1)\).

(b)

Here \(a_n = \dfrac {(x-2)^n}{n^2}\), so \[\left |\frac {a_{n+1}}{a_n}\right | = \frac {n^2}{(n+1)^2}\left |x-2\right | \longrightarrow \left |x-2\right | ,\] giving convergence for \(\left |x-2\right |<1\), that is \(1<x<3\). At both endpoints the absolute series is \(\sum \dfrac {1}{n^2}\), which converges. The interval is therefore the closed interval \([1,3]\).

Solution.

(a)

The derivatives of \(\sin \) cycle through \(\cos , -\sin , -\cos , \sin \), taking the values \(1,0,-1,0\) at the origin, so \[\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \cdots = \sum _{k=0}^{\infty }\frac {(-1)^kx^{2k+1}}{(2k+1)!}.\] Every derivative is bounded by \(1\), so the remainder tends to \(0\) for every \(x\): the interval is \(-\infty <x<\infty \).

(b)

Dividing the series for \(\sin x\) by that for \(\cos x\), \[\tan x = x + \frac {x^3}{3} + \frac {2x^5}{15} + \cdots \] Here \(\tan \) has poles at \(\pm \dfrac {\pi }{2}\), so the radius of convergence is the distance from \(0\) to the nearest one: the interval is \(-\dfrac {\pi }{2}<x<\dfrac {\pi }{2}\).

(c)

Integrating the geometric series \(\dfrac {1}{1+x} = 1-x+x^2-\cdots \) term by term, \[\ln (1+x) = x - \frac {x^2}{2} + \frac {x^3}{3} - \cdots = \sum _{n=1}^{\infty }\frac {(-1)^{n-1}x^n}{n}.\] The ratio test gives \(\left |x\right |<1\). At \(x=1\) the alternating harmonic series converges; at \(x=-1\) it becomes the negative harmonic series and diverges. The interval is \(-1<x\leq 1\).

(d)

Write the exponent in terms of \(x+2\): \[e^{x/2} = e^{\frac {(x+2)-2}{2}} = e^{-1}e^{\frac {x+2}{2}} = e^{-1}\sum _{n=0}^{\infty }\frac {1}{n!}\Big (\frac {x+2}{2}\Big )^n .\] The exponential series converges everywhere, so the interval is \(-\infty <x<\infty \).

(e)

Write \(x = 2 + (x-2)\) and factor out the \(2\): \[\ln x = \ln 2 + \ln \Big (1+\frac {x-2}{2}\Big ) = \ln 2 + \sum _{n=1}^{\infty }\frac {(-1)^{n-1}(x-2)^n}{n\,2^n},\] using part (c) with argument \(\dfrac {x-2}{2}\). That series requires \(\left |\dfrac {x-2}{2}\right |<1\), that is \(0<x<4\), and the right endpoint \(x=4\) converges as before. The interval is \(0<x\leq 4\), which is as expected since \(\ln x\) fails to exist at \(x=0\).

Solution.

(a)

Rather than differentiate \(\cos ^2x\) repeatedly, use the double-angle identity: \[\cos ^2x = \frac {1+\cos 2x}{2}.\] Substituting the cosine series with argument \(2x\), \[\cos 2x = \sum _{n=0}^{\infty }\frac {(-1)^n(2x)^{2n}}{(2n)!} = 1 + \sum _{n=1}^{\infty }\frac {(-1)^n2^{2n}x^{2n}}{(2n)!},\] so \[\cos ^2x = \frac 12 + \frac 12 + \sum _{n=1}^{\infty } \frac {(-1)^n2^{2n-1}x^{2n}}{(2n)!} = 1 + \sum _{n=1}^{\infty }\frac {(-1)^n2^{2n-1}x^{2n}}{(2n)!},\] which is the stated expansion. Checking the first two terms: \(n=1\) gives \(-\dfrac {2}{2!}x^2 = -x^2\) and \(n=2\) gives \(\dfrac {2^3}{4!}x^4 = \dfrac {x^4}{3}\). Since the cosine series converges for every argument, so does this, for all \(x\).

(b)

There is no neat closed form, so compute the derivatives at \(0\), or equivalently invert the cosine series. Writing \(\sec x = \dfrac {1}{\cos x}\) and using \[\cos x = 1 - \frac {x^2}{2} + \frac {x^4}{24} - \frac {x^6}{720} + \cdots ,\] long division gives \[\sec x = 1 + \frac {x^2}{2} + \frac {5x^4}{24} + \frac {61x^6}{720} + \cdots \] The interval is set by the nearest singularity: \(\cos x\) vanishes first at \(\pm \dfrac {\pi }{2}\), so the expansion is valid on \(-\dfrac {\pi }{2}<x<\dfrac {\pi }{2}\).

Solution.

(a)

Every derivative of \(e^x\) is \(e^x\), taking the value \(e^3\) at \(x=3\). So the Taylor series about \(3\) is \[e^x = \sum _{n=0}^{\infty }\frac {e^3}{n!}(x-3)^n = e^3\sum _{n=0}^{\infty }\frac {(x-3)^n}{n!},\] which is the stated form. It may also be seen at once by writing \(e^x = e^3e^{x-3}\) and expanding the second factor.

Convergence is for all \(x\): the derivatives are bounded on any bounded interval, so the remainder tends to zero.

(b)

The derivatives of \(\cos \) cycle as \(-\sin , -\cos , \sin , \cos \). At \(x = \dfrac {\pi }{4}\) both \(\sin \) and \(\cos \) equal \(\dfrac {1}{\sqrt 2}\), so the values run \[\frac {1}{\sqrt 2},\ -\frac {1}{\sqrt 2},\ -\frac {1}{\sqrt 2}, \ \frac {1}{\sqrt 2},\ \frac {1}{\sqrt 2},\ \dots \] repeating with period \(4\) in the pattern \(+,-,-,+\). Hence \[\cos x = \frac {1}{\sqrt 2}\Big [1 - \Big (x-\frac {\pi }{4}\Big ) - \frac {\big (x-\frac {\pi }{4}\big )^2}{2!} + \frac {\big (x-\frac {\pi }{4}\big )^3}{3!} + \cdots \Big ],\] valid for all \(x\) since every derivative is bounded by \(1\).

Solution. Taylor’s theorem with the Lagrange form of the remainder gives, for some \(\xi \) strictly between \(0\) and \(x\), \[R_n(x) = \frac {f^{(n)}(\xi )}{n!}x^n = \frac {e^{\xi }}{n!}x^n ,\] since every derivative of \(e^x\) is \(e^x\).

The case \(x<0\)

Here \(x<\xi <0\), so \(e^{\xi }<e^0 = 1\) because the exponential is increasing. Therefore \[\left |R_n(x)\right | = \frac {e^{\xi }\left |x\right |^n}{n!} < \frac {\left |x\right |^n}{n!}.\]

The case \(x>0\)

Here \(0<\xi <x\), so \(e^{\xi }<e^{x}\), and since \(x^n>0\), \[R_n(x) = \frac {e^{\xi }x^n}{n!} < \frac {x^ne^{x}}{n!}.\]

In both cases the bound tends to \(0\) as \(n\rightarrow \infty \) for each fixed \(x\), because \(\dfrac {\left |x\right |^n}{n!}\rightarrow 0\). That is precisely what proves the series represents \(e^x\) everywhere, rather than merely converging.

Solution. The Maclaurin series is \[\cos x = 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \cdots = \sum _{k=0}^{\infty }\frac {(-1)^kx^{2k}}{(2k)!} .\] By the ratio test, \[\left |\frac {a_{k+1}}{a_k}\right | = \frac {x^2}{(2k+2)(2k+1)}\longrightarrow 0\] for every \(x\), so the series converges everywhere. It also represents \(\cos x\) everywhere, because every derivative of \(\cos \) is bounded by \(1\) in absolute value, so the remainder satisfies \[\left |R_n(x)\right | \leq \frac {\left |x\right |^{n+1}}{(n+1)!} \longrightarrow 0 \qquad \text {for each fixed }x .\] The interval is therefore all of \(\mathbb {R}\), that is \(-\infty <x<\infty \).

The same argument applies to the Taylor series about any point \(a\): the derivative bound does not depend on where the expansion is centred, so it too represents \(\cos x\) on the whole real line.

Solution. The intrinsic equation relates arc length \(s\) to the tangent angle \(\psi \), where \(\tan \psi = \dfrac {dy}{dx}\).

From \(y = \dfrac {\cosh 4x}{4}\) we get \(\dfrac {dy}{dx} = \sinh 4x\), so \[\tan \psi = \sinh 4x .\] For the arc length, using \(1 + \sinh ^2 = \cosh ^2\), \[s = \int _0^x\sqrt {1 + \sinh ^2 4t}\,dt = \int _0^x\cosh 4t\,dt = \frac {\sinh 4x}{4}.\] Comparing the two displays, \[s = \frac {\tan \psi }{4}, \qquad \text {that is}\qquad 4s = \tan \psi .\]

Solution. \[\frac {dy}{dx} = \frac {\sec x\tan x}{\sec x} = \tan x \implies \tan \psi = \tan x \implies \psi = x .\] The tangent angle is the abscissa itself. For the arc length, \[s = \int _0^x\sqrt {1+\tan ^2 t}\,dt = \int _0^x\sec t\,dt = \Big [\ln \left |\sec t + \tan t\right |\Big ]_0^x = \ln \left |\sec x + \tan x\right | .\] Substituting \(\psi = x\), \[s = \ln \left |\sec \psi + \tan \psi \right | .\]

Solution. \[\frac {dx}{dt} = 2 + 2\cos t, \qquad \frac {dy}{dt} = 2\sin t ,\] so, using the half-angle identities \(1+\cos t = 2\cos ^2\dfrac t2\) and \(\sin t = 2\sin \dfrac t2\cos \dfrac t2\), \[\tan \psi = \frac {dy}{dx} = \frac {2\sin t}{2+2\cos t} = \frac {2\sin \frac t2\cos \frac t2}{2\cos ^2\frac t2} = \tan \frac {t}{2} \implies \psi = \frac {t}{2}.\] For the arc length, \[\frac {ds}{dt} = \sqrt {(2+2\cos t)^2 + 4\sin ^2 t} = \sqrt {8 + 8\cos t} = 4\cos \frac {t}{2},\] so \[s = \int _0^t 4\cos \frac {u}{2}\,du = 8\sin \frac {t}{2} .\] Since \(\psi = \dfrac t2\), the intrinsic equation is \[s = 8\sin \psi .\]

Solution. \[\frac {dy}{dx} = 2x-1, \qquad \frac {d^2y}{dx^2} = 2 .\] At \(x = \dfrac 78\), \(\dfrac {dy}{dx} = \dfrac 74 - 1 = \dfrac 34\). Hence \[K = \frac {\left |y''\right |}{\big (1+(y')^2\big )^{3/2}} = \frac {2}{\Big (1+\frac {9}{16}\Big )^{3/2}} = \frac {2}{\Big (\frac {25}{16}\Big )^{3/2}} = \frac {2}{\frac {125}{64}} = \frac {128}{125},\] and the radius of curvature is \[\rho = \frac {1}{K} = \frac {125}{128}.\]

Solution. For a parametrised curve, \[K = \frac {\left |\dot x\ddot y - \dot y\ddot x\right |} {\big (\dot x^2 + \dot y^2\big )^{3/2}} .\] Here \[\dot x = 2\sinh t - 1,\quad \ddot x = 2\cosh t,\qquad \dot y = \sinh t + 1,\quad \ddot y = \cosh t .\] At \(t = \ln 2\), \(\cosh t = \dfrac 54\) and \(\sinh t = \dfrac 34\), so \[\dot x = \frac 12,\quad \dot y = \frac 74,\quad \ddot x = \frac 52,\quad \ddot y = \frac 54 .\] Then \[\dot x\ddot y - \dot y\ddot x = \frac 12\cdot \frac 54 - \frac 74\cdot \frac 52 = \frac 58 - \frac {35}{8} = -\frac {30}{8} = -\frac {15}{4},\] \[\dot x^2 + \dot y^2 = \frac 14 + \frac {49}{16} = \frac {53}{16}.\] Hence \[K = \frac {\frac {15}{4}}{\Big (\frac {53}{16}\Big )^{3/2}} = \frac {\frac {15}{4}\cdot 64}{53\sqrt {53}} = \frac {240}{53\sqrt {53}} = \frac {240\sqrt {53}}{2809}\approx 0.622 .\]

Solution.

Maximum curvature of \(y = \frac 13x^3\)

\(y' = x^2\) and \(y'' = 2x\), so \[K(x) = \frac {2\left |x\right |}{\big (1+x^4\big )^{3/2}} .\] Taking \(x>0\) and differentiating, \[K'(x) = \frac {2\big (1+x^4\big )^{3/2} - 2x\cdot \frac 32\big (1+x^4\big )^{1/2}4x^3} {\big (1+x^4\big )^3} = \frac {2\big (1 - 5x^4\big )}{\big (1+x^4\big )^{5/2}} ,\] which vanishes when \(x^4 = \dfrac 15\), that is \[x = \frac {1}{\sqrt [4]{5}} = \frac {5^{3/4}}{5}\approx 0.669 .\] The corresponding point is \(\Big (5^{-1/4},\ \dfrac {5^{-3/4}}{3}\Big )\), and by symmetry \(x = -5^{-1/4}\) gives the other maximum. There \(K = \dfrac {5\sqrt [4]{5}\sqrt 6}{18}\approx 1.017\).

Circle of curvature of \(y^2 = 12x\) at \((3,6)\)

Taking \(y = \sqrt {12x}\), \(y' = \dfrac {6}{\sqrt {12x}}\) and at \(x=3\), \(y' = \dfrac {6}{6} = 1\); differentiating again, \(y'' = -\dfrac 16\) at that point. Hence \[\rho = \frac {\big (1+(y')^2\big )^{3/2}}{\left |y''\right |} = \frac {2^{3/2}}{\frac 16} = 12\sqrt 2 .\] The centre of curvature is \[\Big (x - \frac {y'\big (1+(y')^2\big )}{y''},\ y + \frac {1+(y')^2}{y''}\Big ) = \Big (3 - \frac {1\cdot 2}{-\frac 16},\ 6 + \frac {2}{-\frac 16}\Big ) = (15, -6).\] As a check, the distance from \((15,-6)\) to \((3,6)\) is \(\sqrt {12^2+12^2} = 12\sqrt 2 = \rho \). The circle of curvature is therefore \[(x-15)^2 + (y+6)^2 = 288 .\]