Approximate expression under square root

Question

Given $x>>1,$ how can the expression \begin{equation} \left(1-\frac{1}{4x^2}\right)^{1/2} \end{equation} be approximated to \begin{equation} \left(1-\frac{1}{8x^2}\right)? \end{equation}

Answer 1

$$\begin{align} 1-\frac{1}{4x^2} &\approx 1-\frac{1}{4x^2}+\frac{1}{64x^4}\\ &= \left (1-\frac{1}{8x^2}\right)^2 \end{align}$$

Answer 2

Let $y=1-1/4x^2$. Then $(1-1/8x^2)^2=y+1/64x^4=y(1+1/64x^4y)>y.$ So$$ y^{1/2}<1-1/8x^2=y^{1/2}(1+1/64x^4y)^{1/2}<y^{1/2}(1+1/128x^4 y).$$ Even for the weak restriction $x\geq 1$ we have $1>y\geq 3/4, $ so the proportionate error is less than $1/128x^4y,$ which is at most $1/ 96 x^4$.

Also $y^{1/2}<1, $ and hence for the value of the error, if $x\geq 1$ we have $$0<y^{1/2}-(1-1/8x^2)<y^{1/2}(1/128z^4 y)\leq y^{1/2}(1/96 x^4)<1/96x^4.$$