# How to find a function's approximation?

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 $$f(x).$$

• I have typed out your question - please refrain from posting images in the future as not all users can see them – lioness99a Jan 25 at 13:41
• Thank you a lot :) – Jack Hwo Jan 25 at 13:50

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

• That is not a linear approximation ... lol – Jack Hwo Jan 25 at 13:53
• I have presented my idea to you, one which you can continue. It's not ready yet. – Matti P. Jan 25 at 13:54
• Okay, thank you :) – Jack Hwo Jan 25 at 13:56

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.

• In this problem, we are looking at $x \approx 0$, not $\frac{1}{2+x}\approx 0$. I also almost fell into this trap ... – Matti P. Jan 25 at 13:58
• Yep, that's the answer the solution offers. But I wonder why... – Jack Hwo Jan 25 at 14:01
• The approximation $(1 + y)^k \approx 1+yk$ holds when $y$ is near zero; you are applying it in a case where $y = -\frac1{2+x}$ is near $-1/2$. – Mees de Vries Jan 25 at 14:03
• Thank you, I am now able to understand the problem. I think the question doesn't make sense a lot. Next time, I'd rather use derivatives to figure out the approximation. Any way, thank you:) – Jack Hwo Jan 25 at 14:14

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 $$0.0 \leq x \leq 0.2$$.