Binomial Expansion

Question

If $f(x)= \dfrac{1}{(2x+1)} + \dfrac{4}{(x-2)} + \dfrac{8}{(x-2)^2}$ show that when $x$ is sufficiently small for $x^3$ and higher powers to be neglected, $f(x)= 1- x + 5x^2$ How do I do this without using the Maclaurin expansion.

Answer 1

HINT : you have $x$ sufficiently small ie you can use the following two expansions :

$$\frac{1}{1-x} = 1+ x + x^2 + O(x^3)$$

$$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + O(x^3)$$

Make sure you transform what you have in the above forms before expanding.

EDIT : If you really want to use the binomial theorem which is : $$(c+y)^\alpha = c^\alpha + \alpha y c^{\alpha-1} + \frac12\alpha(\alpha-1)y^2 c^{\alpha-2} + O(y^3)$$

Then go ahead substituting $c = 1$, $y = -x$ and $\alpha = -1$ and you will get the above geometric series. That said you can just use it directly without the need of using my hints above.