Taylor series expansion limitation 
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*Appplying taylor to any functions its not necessary that that polynomial which gets formed from taylor will work for all x , taylor doesnt gives a condition for when the series is always equal to the function right ?


*We would need some other methods to check whether the series always matches with the function right ? Like in case of ln(1+x) where expanding using taylor around x=0 we will get a series which is only valid for |x|<=1 , or is it that we can have a series around some number like x= 5 for ln(1+x) then the series will always converge ? If yes when and when not to check for series convergence ?


*Is there a deep reason as to why taylor fails to give the whole function in terms of polynomial when expanded around x= 0 but ut might work when expanded around some other point ?
 A: Surprisingly, the answers to your questions require some complex analysis, even though your questions are ostensibly about real-valued functions of a real variable.
That said:
1:
Taylor's theorem with the Lagrange remainder says that if $\lim_{n \to \infty} \frac{|f^{(n)}(x_1)|}{n!} (x-x_0)^n = 0$ whenever $x \in (a,b)$ and $x_1$ is between $x$ and $x_0$ then the Taylor series centered at $x_0 \in (a,b)$ converges to $f$ at all points of $(a,b)$. However this result can be extremely hard to check.
2:
Complex analysis tells us that the radius of convergence of the Taylor series at a given $x_0$ is the distance from $x_0$ to the nearest complex singularity. For $\operatorname{Log}(z)$ (the principal complex logarithm) for example this is the distance from $x_0$ to $0$, which is a singularity of $\operatorname{Log}$ because it is a branch point. Thus for instance $\ln(1+x)$ centered at $x_0=5$ has radius of convergence $6$.
Complex analysis also tells us that if $f$ is complex differentiable on a disk of radius $r$ centered at $z_0$, then the Taylor series of $f$ at $z_0$ converges to $f$ on a disk of radius $R=\min \left \{ r,\left ( \limsup_{n \to \infty} \left ( \frac{|f^{(n)}(z_0)|}{n!} \right )^{1/n}  \right )^{-1} \right \}$, where $0^{-1}$ is understood as $+\infty$. We can't freely drop this "if $f$ is complex differentiable..." assumption. If we do drop it, we can still say the series converges on a disk of radius $\left ( \limsup_{n \to \infty} \left ( \frac{|f^{(n)}(z_0)|}{n!} \right )^{1/n} \right )^{-1}$ but we cannot be sure that the limit is actually $f$.
3:
I think this is answered by my answer to #2.
This gets much more counterintuitive with functions that look completely innocuous on the real line, such as $\frac{1}{x^2+1}$. Here the series at $x_0=0$ has radius of convergence $1$ even though the restriction to the real line has no singularities anywhere. In some sense the restriction "knows" about the singularity at $\pm i$.
