Methods I Tripos Questions

This section contains worked examples from past Tripos papers relevant to Methods I topics.

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Limits and Continuity

Tripos 2016, Paper 2, Question 19Y

Calculate for real $x$ and constant $a$:

(i) $\displaystyle\lim_{x \to 0} \frac{\tan x \sin x}{\sin^3 x}$

(ii) $\displaystyle\lim_{x \to a} \frac{\sin x - \sin a}{x - a}$

(iii) $\displaystyle\lim_{x \to \infty} \left(\frac{x + a}{x - a}\right)^{x+a}$

Solutions:

(i) Using $\tan x = \frac{\sin x}{\cos x}$:

$\lim_{x \to 0} \frac{\sin^2 x}{\cos x \sin^3 x} = \lim_{x \to 0} \frac{1}{\cos x \sin x}$

This doesn’t exist (goes to $\infty$). Actually, let’s reconsider:

$\frac{\tan x \sin x}{\sin^3 x} = \frac{1}{\cos x \sin^2 x}$

Using $\sin x \sim x$ near 0: this behaves like $\frac{1}{x^2}$, so the limit is $\infty$.

(ii) This is the derivative of $\sin x$ at $x = a$:

$\lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \cos a$

Alternatively, use the identity $\sin x - \sin a = 2\cos\frac{x+a}{2}\sin\frac{x-a}{2}$.

(iii) Let $y = \left(\frac{x+a}{x-a}\right)^{x+a}$. Then:

$\ln y = (x+a)\ln\left(1 + \frac{2a}{x-a}\right)$

For large $x$: $\ln\left(1 + \frac{2a}{x-a}\right) \sim \frac{2a}{x-a}$

So $\ln y \sim (x+a) \cdot \frac{2a}{x-a} \to 2a$ as $x \to \infty$.

Therefore: $y \to e^{2a}$

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Tripos 2008, Paper 1, Question 19X

A function is defined for integer $n \in \{0, 1, 2, 3, 4\}$:

$\phi(x, n) = \begin{cases} x^n \sin(1/x) & x \neq 0 \\ 0 & x = 0, n \neq 4 \\ 1 & x = 0, n = 4 \end{cases}$

(i) Determine values of $n$ for which $\phi(x,n)$ tends to a limit as $x \to 0$.

(ii) Determine values of $n$ for which $\phi(x,n)$ is continuous at $x = 0$.

(iii) Determine values of $n$ for which $\phi(x,n)$ is differentiable at $x = 0$.

Solutions:

(i) Limit exists if $n > 0$. For $n = 0$: $\sin(1/x)$ oscillates. For $n \geq 1$: $|x^n \sin(1/x)| \leq |x|^n \to 0$.

(ii) For continuity, need $\lim_{x \to 0} \phi = \phi(0)$. The limit is 0 for $n \geq 1$, so we need $\phi(0) = 0$. This holds for $n \in \{1, 2, 3\}$.

(iii) For differentiability:

$\phi'(0) = \lim_{h \to 0} \frac{\phi(h) - \phi(0)}{h} = \lim_{h \to 0} \frac{h^n \sin(1/h)}{h} = \lim_{h \to 0} h^{n-1}\sin(1/h)$

This exists (equals 0) if $n - 1 > 0$, i.e., $n \in \{2, 3\}$.

For $n = 1$: oscillates. For $n = 0$: unbounded.

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Vectors

Tripos 2005, Paper 1, Question 5C

Consider tetrahedron with vertices at $O(0,0,0)$, $A(1,0,0)$, $B(0,1,0)$, $C(0,0,1)$.

(a) Find the cosines of angles between faces.

(b) Find the areas of the four faces.

(c) Find the distance from $O$ to face $ABC$.

Solution outline:

(a) Normal to face $OAB$ is in $z$-direction. Normal to $OBC$ is in $x$-direction. These are perpendicular. Angle between faces $OAB$ and $OAC$: normals are $(0,0,1)$ and $(0,1,0)$, angle is $90°$.

Face $ABC$ has normal $(1,1,1)$ (from cross product). Angle with other faces uses dot product.

(b) Face $OAB$ is right triangle with area $1/2$. Similarly for $OBC$, $OAC$. Face $ABC$ has area $\frac{\sqrt{3}}{2}$.

(c) Equation of face $ABC$: $\mathbf{r} \cdot (1,1,1) = 1$, or $x + y + z = 1$.

Distance from $O$: $d = \frac{1}{\sqrt{3}}$.

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Tripos 2004, Paper 1, Question 1A

Given vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ with $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \neq 0$, define reciprocal basis $(\mathbf{a}^*, \mathbf{b}^*, \mathbf{c}^*)$ by:

$\mathbf{a}^* = \frac{\mathbf{b} \times \mathbf{c}}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}$

with analogous definitions for $\mathbf{b}^*$, $\mathbf{c}^*$.

Show that $\mathbf{a}^* \cdot \mathbf{a} = 1$ and $\mathbf{a}^* \cdot \mathbf{b} = 0$.

Solution:

$\mathbf{a}^* \cdot \mathbf{a} = \frac{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})} = 1$

$\mathbf{a}^* \cdot \mathbf{b} = \frac{(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{b}}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}$

But $(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{b} = [\mathbf{b}, \mathbf{c}, \mathbf{b}] = 0$ (repeated vector in scalar triple product).

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Series

Tripos 2014, Paper 1

Find the expansion of $\ln(1 + x + x^2)$ for:

(i) small $x$

(ii) large $x$

Solution:

(i) $1 + x + x^2 = 1 + x + x^2$. Let $u = x + x^2$, then for small $u$:

$\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$

Substitute back: $\ln(1+x+x^2) = x + x^2 - \frac{(x+x^2)^2}{2} + \cdots$

(ii) For large $x$: $\ln(1+x+x^2) = \ln(x^2) + \ln\left(\frac{1}{x^2} + \frac{1}{x} + 1\right)$

Let $w = \frac{1}{x} + \frac{1}{x^2}$. Then:

$= 2\ln x + w - \frac{w^2}{2} + \cdots$