$\frac{1}{1+x} -1 \approx -x$ is true?

Question

First of all, I'm sorry to ask too simple math question.
But I have little backgroud knowledge of mathmatics so it took so long time to me.
In my lecture note it says $\frac{1}{1+x} -1 \approx -x$.
How this is possible? Is it a kind of taylor approximation? Or it uses the fact $\log(1+x) \approx x$?

Thanks for your help.

Answer 1

Multiplying by $1+x$, $$\frac1{1+x}-1\approx-x\iff-x\approx-x-x^2.$$ The approximation is good when $x\ll x^2$ or $1\ll x$.

Answer 2

This is Taylor approximation at the first order, and it is true for $x$ small enough only. To see it without to many sophisticated tool, you can notice that $f(x)=\dfrac{1}{1+x}-(1-x)=\dfrac{x^2}{1+x}$.

When $x\to 0$, $f(x)\to 0$, so for $x$ close enough to $0$, $f(x)$ will be small, and even negligible compared to $x$. For example, if $x\geq 0$, you have $0\leq f(x)\leq x^2$ and the error term is quadratic in $x$.

Of course, you can have finer approximations by studying the variations of $f$.

Answer 3

It is precisely Taylor series to first (linear) order.

It is valid, i.e. a good approximation, for small $x$, as you can see in the plot:

For larger $x$, it can be arbitrarily bad (e.g. for $x\to-1$, where the original function diverges.