Functions, limits and continuity – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Substitution gives \(\dfrac00\), so factorise the numerator: \(\dfrac{(x-3)(x+3)}{x-3}=x+3\) for \(x\ne3\) (1). The limit is therefore \(3+3=\) 6 (1).
   2. Again we get \(\dfrac00\). Rewrite using the standard limit \(\dfrac{\sin u}u\to1\): \(\dfrac{\sin2x}{\sin3x}=\dfrac{2}{3}\cdot\dfrac{\sin2x}{2x}\cdot\dfrac{3x}{\sin3x}\) (1). As \(x\to0\) both fractions tend to \(1\), so the limit is \(\dfrac23\) (1).

   In (a), cancelling \(x-3\) is valid because a limit never uses the value at \(x=3\) itself. In (b), a quick check with \(x=0.01\) gives \(\sin0.02\div\sin0.03\approx0.6667\).

2. Answer to Question 2 (4 marks)

   Rough work: \(|(2x+1)-9|=|2x-8|=2|x-4|\), so it is enough to make \(2|x-4|\lt \varepsilon\) (1). Proof: let \(\varepsilon>0\) be given and choose \(\delta=\dfrac\varepsilon2>0\) (1). Suppose \(0\lt |x-4|\lt \delta\) (1). Then \(|(2x+1)-9|=2|x-4|\lt 2\delta=\varepsilon\) (1). Since this works for every \(\varepsilon>0\), the definition is satisfied and \(\displaystyle\lim_{x\to4}(2x+1)=9\).

   The order matters: \(\delta\) is chosen after \(\varepsilon\) and may depend on it. Any smaller positive \(\delta\), such as \(\tfrac\varepsilon3\), also works. A common error is to assume what must be proved, starting from \(|(2x+1)-9|\lt\varepsilon\).

3. Answer to Question 3 (4 marks)

   1. The left-hand limit is \(\displaystyle\lim_{x\to1^-}(x^2+a)=1+a\) (1). The right-hand value is \(f(1)=3-2=1\) (1). Continuity requires these to be equal, so \(1+a=1\) and \(a=0\) (1).
   2. With \(a=0\), the left piece \(x^2\) has gradient \(2x=2\) at \(x=1\), while the right piece \(3x-2\) has gradient \(3\) (1). The one-sided derivatives differ, so \(f\) is not differentiable at 1: the graph has a corner there.

4. Answer to Question 4 (4 marks)

   1. \(f\) is a polynomial, so it is continuous on \([1,2]\). We have \(f(1)=1+2-6=-3\lt 0\) and \(f(2)=8+4-6=6>0\) (1). Since \(f\) changes sign, the intermediate value theorem gives a root in \((1,2)\) (1).
   2. Evaluate at the midpoint: \(f(1.5)=3.375+3-6=0.375>0\) (1). The sign change is now between \(1\) and \(1.5\), so the root lies in \((1,1.5)\) (1). The interval has been halved; repeating the step would locate the root more precisely.

5. Answer to Question 5 (4 marks)

   For every \(x\ne0\), \(-1\le\cos\frac1x\le1\). Multiplying by \(x^2\ge0\) keeps the inequalities, so \(-x^2\le x^2\cos\frac1x\le x^2\) (2). Both bounds tend to \(0\) as \(x\to0\), so by the squeeze theorem the limit is \(0\) (1). On the other hand, \(\cos\frac1x\) takes the value \(1\) at \(x=\frac1{2n\pi}\) and \(-1\) at \(x=\frac1{(2n+1)\pi}\); both sequences tend to \(0\) but give different values, so no limit exists (1).

Total : 20 marks