Expansion of $(a + b x^4 + c x^6 + dx^8)^{-1} $

Question

I am asked to expand $$ (a + b x^4 + c x^6 + dx^8)^{-1} $$ and take the limit $x \rightarrow 0$ of the result. Is it fair neglect the higher powers at the first place and talk about $$ (a + b x^4 + c x^6 + dx^8)^{-1} \approx (a + b x^4)^{-1}? $$ Here, $a,b,c,d$ are constants.

Answer 1

Yes, here it's fine to neglect the higher powers since we are taking the limit as $x\rightarrow 0,$ so terms of order $O(x)$ and higher vanish.

For $a\neq 0$, we should obtain

$$\lim_{x\rightarrow 0} \space (a+bx^4+cx^6+dx^8)^{-1}=\frac{1}{a}$$

We have $$\frac{1}{(a+bx^4+cx^6+dx^8)}=\frac{1}{a}\frac{1}{1-(-\frac{1}{a}(bx^4+cx^6+dx^8))}$$ $$=\frac{1}{a}\big(1-\frac{1}{a}(bx^4+cx^6+dx^8)+\frac{1}{a^2}(bx^4+cx^6+dx^8)^2+...)$$ $$=\frac{1}{a}-\frac{b}{a^2}x^4+...=\frac{1}{a}+O(x^4)\space \underbrace{\rightarrow}_{x\rightarrow 0}\space\frac{1}{a}$$