I have the following function
$$f = \frac{a^2bc}{(\sqrt{b^2c^2 + abc\delta}- bc)^2}$$
and am given that $0 < \delta \ll \frac{bc}{a}$ and $a, b, c > 0$. This motivates me to consider the Taylor expansion of the term under the square root to obtain
$$\sqrt{b^2c^2 + abc\delta} = bc\left(1 + \frac{a\delta}{2bc} + O\left(\frac{a\delta}{bc}\right)^2+\ ... \right)$$
Can I ignore these higher order terms and claim the following?
$$f \approx \frac{4bc}{\delta}$$
If so, what is the error in $f$ that I make due to this approximation?
You are considering $$f = \frac{a^2bc}{(\sqrt{b^2c^2 + abc\delta}- bc)^2}$$ where $0 < \delta \ll \frac{bc}{a}$.
Let $\delta=k \frac{bc}{a}$ to make $$f=\frac{a^2}{b c} \frac{1}{ \left(\sqrt{k+1}-1\right)^2}$$ Now, develop as Laurent series for small $k$ to get $$f=\frac{a^2}{b c} \left(\frac{4}{k^2}+\frac{2}{k}-\frac{1}{4}+\frac{k}{8}+O\left(k^2\right) \right)$$ Back to $\delta$
$$f=\frac{4 b c}{\delta ^2}+\frac{2 a}{\delta }-\frac{a^2}{4 b c}+\frac{a^3 \delta }{8 b^2 c^2}+\cdots$$