Maclaurin approximation conundrum This is the Maclaurin approximation for sinx up to 4 terms. I had this preconceived notion that maclaurin  approximation is used to approximate the function "at x=0". This graph however, seems to suggest more like "around x=0" where the definition of "around" just keeps on increasing on increasing the no. Of terms in my approximation. Does this suggest that if on taking infinite no. Of terms, I'll probably see the entire function approximated or would this fine degree of approximation relent somewhere. If the latter is true, then where would that be on the x-axis? 

 A: If you take all the infinite number of terms, you'll have exactly $\sin x$ on the whole real line.  There's a remainder term for the Maclaurin series which tells you about how big your error is when you use only finitely many terms.  The error gets smaller for larger $n$, but bigger for larger (positive and negative) $x$.
Bonus points for the unintended pun on "entire".
A: McLaurin does not approximate a function "at", because it reads
$$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\cdots$$ and you need to know $f(a)$ beforehand (as well as all derivatives). So it does approximate "around".

The more terms you add, the closer you get to the function, provided the latter is smooth enough (if there is a sudden jump say at $x=1$, the behavior at $x=0$ cannot predict it). 
For many functions, the Taylor development only converges in a finite interval around the "at" point. For example,
$$\frac1{1-x}=1+x+x^3+x^4+\cdots$$
works only for $-1<x<1$.
A: Typically, Taylor expansion of succcessively higher orders is giving you better and better approximations in a small neighborhood around the point of expansion. The neighborhood may shrink or grow with the order depending on the function you are dealing with. For analytic functions like polynomials, $\sin$, or $\exp$, it turns out to grow, but for others it turns out to shrink. 
It would not make much sense to say that a Taylor polynomial approximates the value of the function at the point of expansion because it actually exactly agrees with the function at the point of expansion.
