0
$\begingroup$

How can this function be expanded by simplifying and using binomial expansion to obtain a linear approximation (i.e. the first two terms of the expansion)? $$\frac{3+x-x^2}{3-x+x^2}$$

1 Answers 1

0

Hint:

$$\frac{1+t}{1-t}=1+\frac{2t}{1-t}=1+2t(1+t+t^2+t^3+\cdots).$$

Substitute $t$ with $\dfrac{x-x^2}3$.