Unit 26 Review Exercises
Limits
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. AHL marks questions for HL students only.
1 Paper 1 Let \(f(x) = \dfrac{x^2 + 2x - 8}{x - 2}\).
(a) Find \(f(1.99)\) and \(f(2.01)\).
(b) Hence estimate \(\displaystyle\lim_{x \to 2} f(x)\).
For \(x \neq 2\), \(f(x) = \dfrac{(x + 4)(x - 2)}{x - 2} = x + 4\).
(a) \(f(1.99) = 5.99\), \(\;f(2.01) = 6.01\).
(b) Both sides approach \(6\), so \(\displaystyle\lim_{x \to 2} f(x) = 6\). See 24.1.
2 Paper 1 AHL Find \(\displaystyle\lim_{x \to -2} \frac{x^2 - 4}{x^3 + 8}\).
\(x^3 + 8 = (x + 2)(x^2 - 2x + 4)\), so
\[ \lim_{x \to -2} \frac{(x + 2)(x - 2)}{(x + 2)(x^2 - 2x + 4)} = \frac{-4}{4 + 4 + 4} = -\frac{1}{3}. \]
See 26.1.
3 Paper 1 AHL Find \(\displaystyle\lim_{x \to 0} \frac{\sin 2x \tan 3x}{x^2}\).
\[ \frac{\sin 2x \tan 3x}{x^2} = 2 \times \frac{\sin 2x}{2x} \times 3 \times \frac{\tan 3x}{3x} \to 2 \times 3 = 6. \]
See 26.2.
4 Paper 1 AHL Find \(\displaystyle\lim_{x \to \infty} \frac{6x^3 - x}{2x^3 + 5x^2 + 1}\), and state the equation of the horizontal asymptote of \(y = \dfrac{6x^3 - x}{2x^3 + 5x^2 + 1}\).
Divide by \(x^3\): \(\dfrac{6 - \frac{1}{x^2}}{2 + \frac{5}{x} + \frac{1}{x^3}} \to 3\). The horizontal asymptote is \(y = 3\). See 26.2.
5 Paper 1 AHL Find \(\displaystyle\lim_{x \to \infty} \left(\sqrt{4x^2 + 12x} - 2x\right)\).
\[ \frac{(4x^2 + 12x) - 4x^2}{\sqrt{4x^2 + 12x} + 2x} = \frac{12}{\sqrt{4 + \frac{12}{x}} + 2} \to \frac{12}{4} = 3. \]
See 26.2.
6 Paper 1 AHL Let \(f(x) = \dfrac{x^2 - 25}{x - 5}\) for \(x \neq 5\), and \(f(5) = p\).
(a) Find \(p\) so that \(f\) is continuous at \(x = 5\).
(b) If \(p = 0\), what kind of discontinuity does \(f\) have at \(x = 5\)?
(a) \(\displaystyle\lim_{x \to 5} (x + 5) = 10\), so \(p = 10\).
(b) The limit exists (\(10\)) but is not equal to \(f(5) = 0\): a removable discontinuity (a hole). See 26.3.
7 Paper 1 AHL Let \(f(x) = 3x^2 - x + 2\).
(a) Use first principles to find \(f'(x)\).
(b) Find the gradient of the tangent at \(x = 1\).
(a) \(f(x + h) - f(x) = 3(2xh + h^2) - h = 6xh + 3h^2 - h\). \(\;f'(x) = \displaystyle\lim_{h \to 0} (6x + 3h - 1) = 6x - 1\).
(b) \(f'(1) = 5\). See 26.3.
8 Paper 1 AHL Find \(\displaystyle\lim_{x \to 0} \frac{\mathrm{e}^{3x} - 1 - 3x}{x^2}\).
Form \(\frac{0}{0}\) twice: \(\displaystyle\lim_{x \to 0} \frac{3\mathrm{e}^{3x} - 3}{2x} = \lim_{x \to 0} \frac{9\mathrm{e}^{3x}}{2} = \frac{9}{2}\). See 26.4.
9 Paper 1 AHL Find \(\displaystyle\lim_{x \to \infty} \frac{(\ln x)^2}{x}\).
Form \(\frac{\infty}{\infty}\) twice: \(\displaystyle\lim_{x \to \infty} \frac{\frac{2\ln x}{x}}{1} = \lim_{x \to \infty} \frac{2\ln x}{x} = \lim_{x \to \infty} \frac{\frac{2}{x}}{1} = 0\). See 26.4.
10 Paper 2 Let \(g(x) = \dfrac{2^x - 1}{x}\).
(a) Use your GDC to find \(g(x)\), to 4 decimal places, for \(x = \pm 0.01\) and \(x = \pm 0.001\), and hence estimate \(\displaystyle\lim_{x \to 0} g(x)\) to 3 significant figures.
(b) AHL Use l’Hôpital’s rule to find the exact value of the limit.