How to find a function's approximation?

Question

I am having problems with the following question:

Use the linear approximation $(1+x)^k\approx 1+kx$ to find an approximation for the function $f(x)$ for values of $x$ near zero $$f(x)=\sqrt[3]{\left(1-\frac1{2+x}\right)^2}.$$

When $x$ approaches $0$, the fraction part approaches $\frac12$, which is far from $0$. So I wonder how I can apply the approximation formula given by the question to evaluate the linear approximation of $f(x).$

Answer 1

$$ f(x) = \left( 1- \frac{1}{2+x}\right)^{2/3} = \left( 1- \frac{1}{2} \frac{1}{1+x/2}\right)^{2/3} = \left( 1- \frac{1}{2} (1+x/2)^{-1}\right)^{2/3} $$ and $(1+x/2)^{-1} \approx 1-x/2$ so therefore $$ f(x) \approx \left( 1- \frac{1}{2}(1-x/2)\right)^{2/3} $$ ... and so on. Maybe like this?

Answer 2

Maybe like this? $$ f(x) = \left( 1 - \frac 1 {2+x}\right)^{2/3} \approx 1 - \frac 2 {3(2+x)} $$

Edit: indeed, this works when $\frac 1 {2+x}$ is small, which isn't the case when $x$ is close to zero.

Answer 3

The simplest linear model could come from composition of Taylor series built at $x0$. You have $$1-\frac{1}{x+2}=\frac{1}{2}+\frac{x}{4}-\frac{x^2}{8}+O\left(x^3\right)$$ Now, still using Taylor or binomial expansion $$\left(1-\frac{1}{x+2}\right)^{2/3}=\frac{1}{2^{2/3}}+\frac{x}{3\ 2^{2/3}}-\frac{7 x^2}{36\ 2^{2/3}}+O\left(x^3\right)=\frac{1}{2^{2/3}}+\frac{x}{3\ 2^{2/3}}+O\left(x^2\right)$$ which is not bad for $0.0 \leq x \leq 0.2$.