1.1. Given that $x$ is small such that ${x}^{3}$ and higher powers of $x$ can be ignored show that

        ${(1-\frac{1}{3}x)}^{-1}{(2-x)}^{-2}\approx\frac{1}{4}+\frac{1}{3}x+\frac{43}{144}{x}^{2}$

1.2. For which values of $x$ is the approximation in part (a) valid?

1.3. (i) Use your calculator to find the exact fraction of ${(1-\frac{1}{3}x)}^{-1}$ ${(2-x)}^{-2}$  when $x=0.5$

(ii) Use your calculator to find the fraction from the approximation $\frac{1}{4}+\frac{1}{3}x+\frac{43}{144}{x}^{2}$ when $x=0.5$

(iii) Find the percentage error in the approximation, giving your answer to two decimal places.