Unit 26 Exam Practice
Limits
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
This whole unit is HL only AHL 5.12 AHL 5.13, so there is no SL set.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
HL questions
These questions use limits, continuity, first principles and l’Hôpital’s rule, and combine them with differentiation (Units 24–25) and integration (Units 27–28).
HL 1 Paper 1
(a) It is given that \[ \lim_{x \to 2} \frac{x^{2} + ax + b}{x - 2} = 7, \] where \(a\) and \(b\) are constants. Find the value of \(a\) and the value of \(b\). [4]
(b) Find \(\displaystyle\lim_{x \to \infty} \frac{\ln\left(x^{2} + 1\right)}{\ln x}\). [3]
Total 7 marks
(a)
the denominator tends to \(0\), so for the limit to exist the numerator must also tend to \(0\) R1
\(4 + 2a + b = 0\), so \(b = -4 - 2a\) and \(x^{2} + ax - 4 - 2a = (x - 2)(x + 2 + a)\) M1
the limit is \(\displaystyle\lim_{x \to 2} (x + 2 + a) = 4 + a = 7\) A1
\(a = 3\), \(b = -10\) A1
Accept l’Hôpital’s rule for the M1: \(\displaystyle\lim_{x \to 2} \dfrac{2x + a}{1} = 4 + a\).
[4 marks]
(b)
form \(\dfrac{\infty}{\infty}\), so use l’Hôpital’s rule: \(\displaystyle\lim_{x \to \infty} \dfrac{\frac{2x}{x^{2} + 1}}{\frac{1}{x}}\) M1
\(= \displaystyle\lim_{x \to \infty} \dfrac{2x^{2}}{x^{2} + 1} = \lim_{x \to \infty} \dfrac{2}{1 + \frac{1}{x^{2}}}\) A1
\(= 2\) A1
[3 marks]
HL 2 Paper 1
(a) Find \(\displaystyle\lim_{x \to 0} \frac{\sin 3x\,(1 - \cos 2x)}{x^{2}\tan 4x}\). [4]
(b) Find \(\displaystyle\lim_{x \to -\infty} \left(\sqrt{x^{2} + 4x} + x\right)\). [4]
Total 8 marks
(a)
using \(1 - \cos 2x = 2\sin^{2} x\) to rewrite the numerator M1
\(\dfrac{2\sin 3x\sin^{2} x}{x^{2}\tan 4x} = 2 \times \dfrac{\sin 3x}{x} \times \left(\dfrac{\sin x}{x}\right)^{2} \times \dfrac{x}{\tan 4x}\) A1
\(\dfrac{\sin 3x}{x} = 3 \times \dfrac{\sin 3x}{3x} \to 3\) and \(\dfrac{x}{\tan 4x} = \dfrac{1}{4} \times \dfrac{4x}{\tan 4x} \to \dfrac{1}{4}\) A1
limit \(= 2 \times 3 \times 1^{2} \times \dfrac{1}{4} = \dfrac{3}{2}\) A1
[4 marks]
(b)
multiply by \(\dfrac{\sqrt{x^{2} + 4x} - x}{\sqrt{x^{2} + 4x} - x}\) M1
\(= \dfrac{(x^{2} + 4x) - x^{2}}{\sqrt{x^{2} + 4x} - x} = \dfrac{4x}{\sqrt{x^{2} + 4x} - x}\) A1
for \(x < 0\), \(\sqrt{x^{2}} = -x\), so \(\sqrt{x^{2} + 4x} = -x\sqrt{1 + \frac{4}{x}}\) R1
\(\dfrac{4x}{-x\sqrt{1 + \frac{4}{x}} - x} = \dfrac{4}{-\sqrt{1 + \frac{4}{x}} - 1} \to \dfrac{4}{-2} = -2\) A1
Do not award the R1 or the final A1 for the answer \(2\) (from \(\sqrt{x^{2}} = x\)).
[4 marks]
See 26.2.
HL 3 Paper 1
Let \(g(x) = ax^{2} + bx\), where \(a\) and \(b\) are constants.
(a) Use first principles to show that \(g'(x) = 2ax + b\). [3]
The function \(f\) is defined by \[ f(x) = \begin{cases} ax^{2} + bx, & x \leqslant 1 \\ 4\sqrt{x} - 1, & x > 1. \end{cases} \]
(b) Given that \(f\) is continuous at \(x = 1\), find an equation connecting \(a\) and \(b\). [2]
(c) Given also that the two pieces of the graph of \(f\) have the same gradient at \(x = 1\), so that \(f\) is differentiable at \(x = 1\), find the value of \(a\) and the value of \(b\). [3]
Total 8 marks
(a)
\(g(x + h) - g(x) = a(x + h)^{2} + b(x + h) - ax^{2} - bx\) M1
\(= 2axh + ah^{2} + bh\), so \(\dfrac{g(x + h) - g(x)}{h} = 2ax + ah + b\) A1
\(g'(x) = \displaystyle\lim_{h \to 0} (2ax + ah + b) = 2ax + b\) A1 AG
[3 marks]
(b)
\(f(1) = a + b\) and \(\displaystyle\lim_{x \to 1^{+}} \left(4\sqrt{x} - 1\right) = 3\) M1
\(a + b = 3\) A1
[2 marks]
(c)
the gradients of the two pieces must be equal at \(x = 1\), using part (a) for the left piece M1
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(4\sqrt{x} - 1\right) = \dfrac{2}{\sqrt{x}}\), so \(2a + b = \dfrac{2}{\sqrt{1}} = 2\) A1
subtracting \(a + b = 3\): \(a = -1\), \(b = 4\) A1
[3 marks]
See 26.3.
HL 4 Paper 1
Let \(f(x) = x^{4}\ln x\), for \(x > 0\).
(a) Show that \(\displaystyle\lim_{x \to 0^{+}} f(x) = 0\). [3]
(b) Find the coordinates of the stationary point of the graph of \(f\), and show that it is a minimum point. [5]
(c) Find \(\displaystyle\lim_{x \to 0^{+}} f'(x)\). [2]
(d) Let \(0 < a < 1\). Show that \[ \int_a^1 x^{4}\ln x \,\mathrm{d}x = -\frac{1}{25} - \frac{a^{5}\ln a}{5} + \frac{a^{5}}{25}. \] [3]
The region \(R\) is enclosed by the graph of \(f\), for \(0 < x \leqslant 1\), and the \(x\)-axis. The area of \(R\) is the limit, as \(a \to 0^{+}\), of the area of the part of \(R\) with \(a \leqslant x \leqslant 1\).
(e) Find the area of \(R\). [3]
Total 16 marks
(a)
\(x^{4}\ln x = \dfrac{\ln x}{x^{-4}}\), form \(\dfrac{-\infty}{\infty}\) M1
l’Hôpital’s rule: \(\displaystyle\lim_{x \to 0^{+}} \dfrac{\frac{1}{x}}{-4x^{-5}} = \lim_{x \to 0^{+}} \left(-\dfrac{x^{4}}{4}\right)\) A1
\(= 0\) A1 AG
[3 marks]
(b)
\(f'(x) = 4x^{3}\ln x + x^{3}\) A1
\(x^{3}(4\ln x + 1) = 0\) and \(x > 0\), so \(\ln x = -\dfrac{1}{4}\) M1
\(x = \mathrm{e}^{-\frac{1}{4}}\) A1
\(y = \mathrm{e}^{-1} \times \left(-\dfrac{1}{4}\right) = -\dfrac{1}{4\mathrm{e}}\) A1
\(f''(x) = 12x^{2}\ln x + 7x^{2}\); at \(x = \mathrm{e}^{-\frac{1}{4}}\), \(f''(x) = x^{2}(-3 + 7) = 4\mathrm{e}^{-\frac{1}{2}} > 0\), so it is a minimum R1
Accept a sign test of \(f'\): \(4\ln x + 1 < 0\) to the left and \(> 0\) to the right.
[5 marks]
(c)
\(4x^{3}\ln x = \dfrac{4\ln x}{x^{-3}}\), form \(\dfrac{-\infty}{\infty}\); l’Hôpital: \(\dfrac{\frac{4}{x}}{-3x^{-4}} = -\dfrac{4x^{3}}{3} \to 0\) M1
\(\displaystyle\lim_{x \to 0^{+}} f'(x) = 0 + 0 = 0\) A1
[2 marks]
(d)
\(u = \ln x\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = x^{4}\): \(\displaystyle\int x^{4}\ln x \,\mathrm{d}x = \dfrac{x^{5}}{5}\ln x - \int \dfrac{x^{4}}{5} \,\mathrm{d}x\) M1
\(= \dfrac{x^{5}}{5}\ln x - \dfrac{x^{5}}{25}\) \((+\,c)\) A1
\(\left(0 - \dfrac{1}{25}\right) - \left(\dfrac{a^{5}\ln a}{5} - \dfrac{a^{5}}{25}\right) = -\dfrac{1}{25} - \dfrac{a^{5}\ln a}{5} + \dfrac{a^{5}}{25}\) A1 AG
[3 marks]
(e)
\(f(x) < 0\) for \(0 < x < 1\), so the area is \(-\displaystyle\int_a^1 x^{4}\ln x \,\mathrm{d}x = \dfrac{1}{25} + \dfrac{a^{5}\ln a}{5} - \dfrac{a^{5}}{25}\) M1
\(a^{5}\ln a = a \times a^{4}\ln a \to 0 \times 0 = 0\) (by part (a)), and \(a^{5} \to 0\) R1
area of \(R = \dfrac{1}{25}\) A1
Accept \(\displaystyle\lim_{a \to 0^{+}} a^{5}\ln a = 0\) shown by l’Hôpital’s rule for the R1.
[3 marks]
HL 5 Paper 2
The function \(f\) is defined by \[ f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x}, & x \neq 0 \\[2mm] k, & x = 0, \end{cases} \] where \(k\) is a constant.
(a) Find \(\displaystyle\lim_{x \to 0} \frac{1 - \cos 2x}{x}\). Hence write down the value of \(k\) for which \(f\) is continuous at \(x = 0\). [3]
For the rest of this question, \(k\) has this value.
(b) Use the definition \(f'(0) = \displaystyle\lim_{h \to 0} \frac{f(h) - f(0)}{h}\) to show that \(f'(0) = 2\). [3]
(c) For \(x \neq 0\):
(i) show that \(f'(x) = \dfrac{2x\sin 2x + \cos 2x - 1}{x^{2}}\) [2]
(ii) hence find \(\displaystyle\lim_{x \to 0} f'(x)\) [3]
(iii) comment on your answers to part (b) and part (c)(ii). [1]
(d) Find the coordinates of the local maximum point of the graph of \(y = f(x)\) for \(0 < x < \pi\). [2]
(e) Find the area of the region enclosed by the graph of \(y = f(x)\) and the \(x\)-axis for \(0 \leqslant x \leqslant \pi\). [2]
Total 16 marks
(a)
\(1 - \cos 2x = 2\sin^{2} x\), so \(f(x) = 2\sin x \times \dfrac{\sin x}{x}\) for \(x \neq 0\) M1
\(\displaystyle\lim_{x \to 0} f(x) = 2 \times 0 \times 1 = 0\) A1
continuity needs \(f(0) = \displaystyle\lim_{x \to 0} f(x)\), so \(k = 0\) A1
Accept l’Hôpital’s rule for the M1: \(\displaystyle\lim_{x \to 0} \dfrac{2\sin 2x}{1} = 0\).
[3 marks]
(b)
\(f(0) = 0\), so \(\dfrac{f(h) - f(0)}{h} = \dfrac{1 - \cos 2h}{h^{2}}\) M1
\(= \dfrac{2\sin^{2} h}{h^{2}} = 2\left(\dfrac{\sin h}{h}\right)^{2}\) A1
\(\to 2 \times 1^{2} = 2\), so \(f'(0) = 2\) A1 AG
Accept l’Hôpital’s rule used twice: \(\dfrac{2\sin 2h}{2h}\), then \(\dfrac{4\cos 2h}{2} \to 2\).
[3 marks]
(c)
(i)
quotient rule: \(f'(x) = \dfrac{x \times 2\sin 2x - (1 - \cos 2x) \times 1}{x^{2}}\) M1 A1
\(= \dfrac{2x\sin 2x + \cos 2x - 1}{x^{2}}\) AG
[2 marks]
(ii)
form \(\dfrac{0}{0}\); the derivative of the numerator is \(2\sin 2x + 4x\cos 2x - 2\sin 2x\) M1
\(\displaystyle\lim_{x \to 0} f'(x) = \lim_{x \to 0} \dfrac{4x\cos 2x}{2x}\) A1
\(= \displaystyle\lim_{x \to 0} 2\cos 2x = 2\) A1
[3 marks]
(iii)
\(\displaystyle\lim_{x \to 0} f'(x) = f'(0)\), so \(f'\) is continuous at \(x = 0\) (the gradient changes smoothly) R1
[1 mark]
(d)
(for example, using Analyze Graph → Maximum on a GDC) \(x = 1.17\) \((1.16556\ldots)\) A1
\(y = 1.45\) \((1.44922\ldots)\) A1
[2 marks]
(e)
\(f(x) \geqslant 0\) for \(0 \leqslant x \leqslant \pi\), so the area is \(\displaystyle\int_0^{\pi} \dfrac{1 - \cos 2x}{x} \,\mathrm{d}x\) M1
\(= 2.44\) \((2.43765\ldots)\) (using the Integral feature on a GDC) A1
[2 marks]