Sign of a function using Taylor approximation If I have a real function $f(x)$, can I study its sign using Taylor approximation? That is, can the Taylor approximation of this function tell me the sign of the function?
I'll make an example
$$f(x)=\mathrm{arctg} (x) - \mathrm{lg} (1+x) \sim_{x \to 0} \frac{x^2}{2}$$
Can I therefore say that,since $\frac{x^2}{2}>0$, then also $f(x)>0$ when $x \to 0$?
Is there any theorem that links the Taylor approximation with the sign of the function?
 A: Remember that a Taylor expansion is around a given point.
So, in your case, close to $x=0$, the function is positive and locall y behaves as $\frac{x^2}2$ . But the function goes to a maximum, which you can show using 
$$\tan^{-1}(x) - \log (1+x) =\frac{x^2}{2}-\frac{2 x^3}{3}+\frac{x^4}{4}+O\left(x^5\right)$$ The derivative will cancel at $x=1$ and then becomes negative. In fact, you would have an $x$ intercept around $x=2$ and the function stays negative above this value.
A: Yes, sufficiently close to $0$ you must have $f > 0$. Observe that Taylor's theorem gives you an estimate for the remainder term ob the Taylor expansion. For any $x$ you can find a $\lambda > 0$ such that
$$f(x) = \frac{x^2}{2} + f^{(3)}(\lambda x) \frac{x^3}{6} = \frac{x^2}{2} \left(1 + f^{(3)}(\lambda x)\frac{x}{3}\right)$$
holds.
Now you just need to choose $\varepsilon > 0$ so small, that $\sup \limits_{y \in [-\varepsilon, \varepsilon]} \left|f^{(3)}(y) \frac{\varepsilon}{3} \right| < \frac{1}{2}$. Then for any $x$ with $|x| \le \varepsilon$ you get the estimate
$$\frac{x^2}{2} \left(1 + f^{(3)}(\lambda x)\frac{x}{3}\right) \ge \frac{x^2}{2} \left(1 - \left|f^{(3)}(\lambda x)\frac{x}{3}\right|\right) \ge \frac{x^2}{4}.$$
You can use similar arguments to show that a function changes sign at $0$ if e.g. $0 = f(0) = f'(0) = f''(0)$ and $f^{(3)} \ne 0$ and $f$ is a $C^4$-function.
