1. Show that, if neither \(a\) nor \(b\) is zero, then \[ax^{n} + bx^{n-1} + \dots + k = ax^{n} (1 + \epsilon_{x}),\] where \(\epsilon_{x}\) is of the first order of smallness when \(x\) is large.

 

2. If \(P(x) = ax^{n} + bx^{n-1} + \dots + k\), and \(a\) is not zero, then as \(x\) increases \(P(x)\) has ultimately the sign of \(a\); and so has \(P(x + \lambda) – P(x)\), where \(\lambda\) is any constant.

 

3. Show that in general \[(ax^{n} + bx^{n-1} + \dots + k)/(Ax^{n} + Bx^{n-1} + \dots + K) = \alpha + (\beta/x) (1 + \epsilon_{x}),\] where \(\alpha = a/A\), \(\beta = (bA – aB)/A^{2}\), and \(\epsilon_{x}\) is of the first order of smallness when \(x\) is large. Indicate any exceptional cases.

 

4. Express \[(ax^{2} + bx + c)/(Ax^{2} + Bx + C)\] in the form \[\alpha + (\beta/x) + (\gamma/x^{2})(1 + \epsilon_{x}),\] where \(\epsilon_{x}\) is of the first order of smallness when \(x\) is large.

 

5. Show that \[\lim_{x\to\infty}\sqrt{x}\{\sqrt{x + a} – \sqrt{x}\} = \tfrac{1}{2} a.\]

[Use the formula \(\sqrt{x + a} – \sqrt{x} = a/\{\sqrt{x + a} + \sqrt{x}\}\).]

 

6. Show that \(\sqrt{x + a} = \sqrt{x} + \frac{1}{2}(a/\sqrt{x}) (1 + \epsilon_{x})\), where \(\epsilon_{x}\) is of the first order of smallness when \(x\) is large.

 

7. Find values of \(\alpha\) and \(\beta\) such that \(\sqrt{a x^{2} + 2bx + c} – \alpha x – \beta\) has the limit zero as \(x \to \infty\); and prove that \(\lim x\{\sqrt{ax^{2} + 2bx + c} – \alpha x – \beta\} = (ac – b^{2})/2a\).

 

8. Evaluate \[\lim_{x \to\infty} x\left\{\sqrt{x^{2} + \sqrt{x^{4} + 1}} – x\sqrt{2}\right\}.\]

 

9. Prove that \((\sec x – \tan x) \to 0\) as \(x \to \frac{1}{2}\pi\).

 

10. Prove that \(\phi(x) = 1 – \cos(1 – \cos x)\) is of the fourth order of smallness when \(x\) is small; and find the limit of \(\phi(x)/x^{4}\) as \(x \to 0\).

 

11. Prove that \(\phi(x) = x\sin(\sin x) – \sin^{2}x\) is of the sixth order of smallness when \(x\) is small; and find the limit of \(\phi(x)/x^{6}\) as \(x \to 0\).

 

12. From a point \(P\) on a radius \(OA\) of a circle, produced beyond the circle, a tangent \(PT\) is drawn to the circle, touching it in \(T\), and \(TN\) is drawn perpendicular to \(OA\). Show that \(NA/AP \to 1\) as \(P\) moves up to \(A\).

 

13. Tangents are drawn to a circular arc at its middle point and its extremities; \(\Delta\) is the area of the triangle formed by the chord of the arc and the two tangents at the extremities, and \(\Delta’\) the area of that formed by the three tangents. Show that \(\Delta/\Delta’ \to 4\) as the length of the arc tends to zero.

 

14. For what values of \(a\) does \(\{a + \sin(1/x)\}/x\) tend to (1) \(\infty\), (2) \(-\infty\), as \(x \to 0\)? [To \(\infty\) if \(a > 1\), to \(-\infty\) if \(a < -1\): the function oscillates if \(-1 \leq a \leq 1\).]

 

15. If \(\phi(x) = 1/q\) when \(x = p/q\), and \(\phi(x) = 0\) when \(x\) is irrational, then \(\phi(x)\) is continuous for all irrational and discontinuous for all rational values of \(x\).

 

16. Show that the function whose graph is drawn in Fig. 32 may be represented by either of the formulae \[1 – x + [x] – [1 – x],\quad 1 – x – \lim_{n\to\infty} (\cos^{2n+1}\pi x).\]

 

17. Show that the function \(\phi(x)\) which is equal to \(0\) when \(x = 0\), to \(\frac{1}{2} – x\) when \(0 < x < \frac{1}{2}\), to \(\frac{1}{2}\) when \(x = \frac{1}{2}\), to \(\frac{3}{2} – x\) when \(\frac{1}{2}< x < 1\), and to \(1\) when \(x = 1\), assumes every value between \(0\) and \(1\) once and once only as \(x\) increases from \(0\) to \(1\), but is discontinuous for \(x = 0\), \(x = \frac{1}{2}\), and \(x = 1\). Show also that the function may be represented by the formula \[\tfrac{1}{2} – x – \tfrac{1}{2}[2x] – \tfrac{1}{2}[1 – 2x].\]

 

18. Let \(\phi(x) = x\) when \(x\) is rational and \(\phi(x) = 1 – x\) when \(x\) is irrational. Show that \(\phi(x)\) assumes every value between \(0\) and \(1\) once and once only as \(x\) increases from \(0\) to \(1\), but is discontinuous for every value of \(x\) except \(x = \frac{1}{2}\).

 

19. As \(x\) increases from \(-\frac{1}{2}\pi\) to \(\frac{1}{2}\pi\), \(y = \sin x\) is continuous and steadily increases, in the stricter sense, from \(-1\) to \(1\). Deduce the existence of a function \(x = \arcsin y\) which is a continuous and steadily increasing function of \(y\) from \(y = -1\) to \(y = 1\).

 

20. Show that the numerically least value of \(\arctan y\) is continuous for all values of \(y\) and increases steadily from \(-\frac{1}{2}\pi\) to \(\frac{1}{2}\pi\) as \(y\) varies through all real values.

 

21. Discuss, on the lines of §§ 108-109, the solution of the equations \[y^{2} – y – x = 0,\quad y^{4} – y^{2} – x^{2} = 0,\quad y^{4} – y^{2} + x^{2} = 0\] in the neighbourhood of \(x = 0\), \(y = 0\).

 

22. If \(ax^{2} + 2bxy + cy^{2} + 2dx + 2ey = 0\) and \(\Delta = 2bde – ae^{2} – cd^{2}\), then one value of \(y\) is given by \(y = \alpha x + \beta x^{2} + (\gamma + \epsilon_{x}) x^{3}\), where \[\alpha = -d/e,\quad \beta = \Delta/2e^{3},\quad \gamma = (cd – be) \Delta/2e^{5},\] and \({\epsilon_{x}}\) is of the first order of smallness when \(x\) is small.

[If \(y – \alpha x = \eta\) then \[-2e\eta = ax^{2} + 2bx(\eta + \alpha x) + c(\eta + \alpha x)^{2} = Ax^{2} + 2Bx \eta + C\eta^{2},\] say. It is evident that \(\eta\) is of the second order of smallness, \(x\eta\) of the third, and \(\eta^{2}\) of the fourth; and \(-2e\eta = Ax^{2} – (AB/e) x^{3}\), the error being of the fourth order.]

 

23. If \(x = ay + by^{2} + cy^{3}\) then one value of \(y\) is given by \[y = \alpha x + \beta x^{2} + (\gamma + \epsilon_{x}) x^{3},\] where \(\alpha = 1/a\), \(\beta = -b/a^{3}\), \(\gamma = (2b^{2} – ac)/a^{5}\), and \(\epsilon_{x}\) is of the first order of smallness when \(x\) is small.

 

24. If \(x = ay + by^{n}\), where \(n\) is an integer greater than unity, then one value of \(y\) is given by \(y = \alpha x + \beta x^{n} + (\gamma + \epsilon_{x}) x^{2n-1}\), where \(\alpha = 1/a\), \(\beta = -b/a^{n+1}\), \(\gamma = nb^{2}/a^{2n+1}\), and \(\epsilon_{x}\) is of the \((n – 1)\)th order of smallness when \(x\) is small.

 

25. Show that the least positive root of the equation \(xy = \sin x\) is a continuous function of \(y\) throughout the interval \({[0, 1]}\), and decreases steadily from \(\pi\) to \(0\) as \(y\) increases from \(0\) to \(1\). [The function is the inverse of \((\sin x)/x\): apply § 109.]

 

26. The least positive root of \(xy = \tan x\) is a continuous function of \(y\) throughout the interval \({[1, \infty)}\), and increases steadily from \(0\) to \(\frac{1}{2}\pi\) as \(y\) increases from \(1\) towards \(\infty\).


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