<s>[INST] Taylor-expansion of numerator and denominator separately To Taylor-expand $f(x) = \frac{\ln{(x+1)}}{\sin{x}}$, can you expand the numerator and denominator separately?
So if $g(x) = \ln{(x+1)}$ and $h(x) = \sin{x}$, then:
$g'(x) = \frac{1}{x+1}$ and $g''(x) = \frac{-1}{(x+1)^2}$
$h'(x) = \cos{x}$ and $h''(x) = -\sin{x}$
So the Taylor-expansions to second degree about 0 is:
$T_{g(x)}(x) = x - \frac{1}{2}x^2$ and $T_{h(x)}(x) = x$, giving the Taylor-expansion for $f$:
$$T_{f(x)}(x) = 1 - \frac{1}{2}x$$
I don't know if this is true, however it doesn't seem to be very inaccurate if $x=0.01$...
 [/INST] $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$
$$f(x) = \frac{\log{(x+1)}}{\sin(x)}=\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right) } { x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)}$$ Now, using the long division
$$f(x)=1-\frac{x}{2}+\frac{x^2}{2}-\frac{x^3}{3}+\frac{11 x^4}{40}+O\left(x^5\right)=1-\frac{x}{2}+\frac{x^2}{2}+O\left(x^3\right)$$
Use always a few more terms when you need to divide later.
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