Taylor series and limitsEdexcel A-Level Further Maths: Subtopic test

1. 1
   The function $f(x)=\ln x$ is expanded as a Taylor series in ascending powers of $(x-2)$.

   (a)

   Find the coefficient of $(x-2)^2$.

   [1 mark]

   • A$-\frac18$
   • B$-\frac14$
   • C$\frac18$
   • D$\frac14$

   (b)

   Find the coefficient of $(x-2)^3$.

   [1 mark]

   • A$\frac14$
   • B$\frac1{12}$
   • C$-\frac1{24}$
   • D$\frac1{24}$

   (c)

   Use the series up to and including the term in $(x-2)^3$ to estimate $\ln2.2$, giving your answer to 4 decimal places.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   The Maclaurin series $\mathrm{e}^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$, $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots$ and $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$ may be used. All limits are as $x\to0$.

   (a)

   Find $\displaystyle\lim_{x\to0}\frac{\sin x-x}{x^3}$.

   [1 mark]

   • A$-\frac16$
   • B$\frac16$
   • C$0$
   • D$-\frac13$

   (b)

   Find $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}$.

   [1 mark]

   • A$1$
   • B$\frac12$
   • C$0$
   • D$-\frac12$

   (c)

   Find $\displaystyle\lim_{x\to0}\frac{\mathrm{e}^{2x^2}-1}{x^2}$.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Let $f(x)=\cos x$.

   (a)

   Find the Taylor series of $f(x)$ in ascending powers of $\left(x-\frac{\pi}{3}\right)$, up to and including the term in $\left(x-\frac{\pi}{3}\right)^3$.

   [3 marks]

   (b)

   (i) Use your series with $x=1.1$ to estimate $\cos1.1$, giving your answer to 5 decimal places.
   (ii) Given that $\cos1.1=0.4535961$ to 7 decimal places, explain why the estimate in (i) is so accurate.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   You may use the series $\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\dots$ and $\mathrm{e}^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$ for small $x$.

   (a)

   (i) Show that $\displaystyle\lim_{x\to0}\frac{x-\arctan x}{x^3}=\frac13$.
   (ii) Find $\displaystyle\lim_{x\to0}\frac{x-\arctan x-\frac{x^3}{3}}{x^5}$.

   [6 marks]

   (b)

   Find $\displaystyle\lim_{x\to0}\frac{x\arctan x-\left(\mathrm{e}^{x^2}-1\right)}{x^4}$.

   [6 marks]

   Total for question 4: 12 marks

End of questions