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1995 Paper 2 Q4
D: 1600.0 B: 1504.3
integration reduction formula integration by parts Wallis integral trigonometric inequality limits curve sketching squeeze theorem

Let \[ u_{n}=\int_{0}^{\frac{1}{2}\pi}\sin^{n}t\,\mathrm{d}t \] for each integer \(n\geqslant0\). By integrating \[ \int_{0}^{\frac{1}{2}\pi}\sin t\sin^{n-1}t\,\mathrm{d}t \] by parts, or otherwise, obtain a formula connecting \(u_{n}\) and \(u_{n-2}\) when \(n\geqslant2\) and deduce that \[ nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2} \] for all \(n\geqslant2\). Deduce that \[ nu_{n}u_{n-1}=\tfrac{1}{2}\pi. \] Sketch graphs of \(\sin^{n}t\) and \(\sin^{n-1}t\), for \(0\leqslant t\leqslant\frac{1}{2}\pi,\) on the same diagram and explain why \(0 < u_{n} < u_{n-1}.\) By using the result of the previous paragraph show that \[ nu_{n}^{2} < \tfrac{1}{2}\pi < nu_{n-1}^{2} \] for all \(n\geqslant1\). Hence show that \[ \left(\frac{n}{n+1}\right)\tfrac{1}{2}\pi < nu_{n}^{2} < \tfrac{1}{2}\pi \] and deduce that \(nu_{n}^{2}\rightarrow\tfrac{1}{2}\pi\) as \(n\rightarrow\infty\).

Solution: \begin{align*} && u_n &= \int_0^{\tfrac12 \pi} \sin^{n} t \, \d t \\ && &= \int_0^{\tfrac12 \pi} \sin t \sin^{n-1} t \, \d t \\ && &= \left [ -\cos t \sin^{n-1} t \right]_0^{\tfrac12 \pi} + \int_0^{\tfrac12 \pi} \cos t (n-1) \sin^{n-2} t \cos t \d t \\ && &= 0 + (n-1)\int_0^{\tfrac12 \pi} \cos^2 t \sin^{n-2} t \d t \\ && &= (n-1) \int_0^{\tfrac12 \pi}(1-\sin^2 t) \sin^{n-2} t \d t \\ && &= (n-1)u_{n-2} - (n-1)u_n \\ \Rightarrow && n u_n &= (n-1)u_{n-2} \\ \end{align*} Mutplying both sides by \(u_{n-1}\) we obtain \(nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}\). Therefore \(nu_nu_{n-1}\) is constant, ie is equal to \(\displaystyle u_1u_0 = \int_0^{\tfrac12 \pi} \sin^{1} t \, \d t \int_0^{\tfrac12 \pi} \sin^{0} t \, \d t = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}\)

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Since \(0 < \sin t < 1\) for \(t \in (0, \tfrac{\pi}{2})\) we must have \(0 < \sin^n t < \sin^{n-1} t\), in particular \(0 < u_n < u_{n-1}\) Therefore \begin{align*} && nu_{n}u_{n-1} &= \tfrac{1}{2}\pi \\ \Rightarrow && nu_n u_n &< \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_{n-1} u_{n-1} &> \tfrac{1}{2}\pi \tag{\(u_n < u_{n-1}\)} \\ \Rightarrow && nu_n^2 &< \tfrac12 \pi < n u_{n-1}^2 \end{align*} However we also have \(\tfrac12 \pi < (n+1)u_n^2\) (by considering the next inequality), so \(\left ( \frac{n}{n+1}\right) \tfrac12 \pi < n u_n^2 < \tfrac12 \pi\) but since as \(n \to \infty\) the right hand bound is constant and the left hand bound tends to \(\tfrac12 \pi\) therefore \(n u_n^2 \to \tfrac12 \pi\)

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1992 Paper 2 Q1
D: 1600.0 B: 1500.0
limits small angle approximation squeeze theorem rational function trigonometric sinc function arctan surds rationalisation

Find the limit, as \(n\rightarrow\infty,\) of each of the following. You should explain your reasoning briefly. \begin{alignat*}{4} \mathbf{(i)\ \ } & \dfrac{n}{n+1}, & \qquad & \mathbf{(ii)\ \ } & \dfrac{5n+1}{n^{2}-3n+4}, & \qquad & \mathbf{(iii)\ \ } & \dfrac{\sin n}{n},\\ \\ \mathbf{(iv)\ \ } & \dfrac{\sin(1/n)}{(1/n)}, & & \mathbf{(v)}\ \ & (\arctan n)^{-1}, & & \mathbf{(vi)\ \ } & \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}. \end{alignat*}

Solution:

  1. \begin{align*} \lim_{n \to \infty} \frac{n}{n+1} &= \lim_{n \to \infty} \left (1 - \frac{1}{n+1} \right ) \\ &\underbrace{=}_{\text{sum of limits}} \lim_{n \to \infty} 1 - \lim_{n \to \infty} \frac{1}{n+1}\\ &= 1 \end{align*}
  2. \begin{align*} \lim_{n \to \infty} \frac{5n+1}{n^2-3n+4} &= \lim_{n \to \infty} \frac{5/n + 1/n^2}{1-3/n+ 4/n^2} \\ &\underbrace{=}_{\text{ratio of limits}} \frac{\displaystyle \lim_{n \to \infty}(5/n + 1/n^2)}{\displaystyle \lim_{n \to \infty}(1-3/n+ 4/n^2)} \\ &= \frac{0}{1} = 0 \end{align*}
  3. \begin{align*} && \lvert \frac{\sin n}{n} \rvert &\leq \frac{1}{n} \quad \quad (n \geq 1) \\ \Rightarrow && \lim_{n \to \infty} \lvert \frac{\sin n}{n} \rvert &\leq \lim_{n \to \infty}\frac{1}{n} \\ &&&= 0\\ \Rightarrow && \lim_{n \to \infty} \frac{\sin n}{n} &= 0 \end{align*}
  4. First note that \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} \to 1\), then \(\frac1n\) is a sequence converging to zero, therefore \(\frac{\sin 1/n}{1/n}\) also must tend to \(1\).
  5. Note that \(\lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2}\) and since \(n\) is a sequence tending to infinity we must have \(\lim_{n \to \infty} \tan^{-1} n = \frac{\pi}{2}\)
  6. \begin{align*} \lim_{n \to \infty} \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}} &= \lim_{n \to \infty} \dfrac{\frac{1}{\sqrt{n+1}+\sqrt{n}}}{\frac{2}{\sqrt{n+2}+\sqrt{n}}} \\ &= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\ &= \frac12 \lim_{n \to \infty} \dfrac{\sqrt{1+2/n}+\sqrt{1}}{\sqrt{1+1/n}+\sqrt{1}}\\ &= \frac12 \end{align*}

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