Describing behaviour of a function around origin Recently, I saw one of the lecturers use Taylor's expansion series to describe the behaviour of $\sinh x$ at the origin and I did not quite understand how come by approximating our initial function ($\sinh x$) we obtain its approximate behavior around origin and not anywhere else. For instance:
$$\sinh x = \frac{e^x - e^{-x}}2 = \frac{(1+x+\frac12x^2+\dots) - (1-x+\frac12x^2-\dots)}2 = x$$
which is indeed true as $\sinh x$ behaves as $y=x$ at the origin.
My question is why this is true? Why approximation of the function gives the behaviour around origin and not anywhere else?
I hope this does not sound confusing and thank you for your help!
 A: Say the effect of higher order terms is bounded by coefficient $B$ we can then consider $$f(x) = x+c_2x^2+\cdots < x + \underset{\text{geometric series}}{\underbrace{B\left(\sum_{k=2}^\infty x^k \right)}}$$
The geometric series is convergent on $[-1,1]$ and we can show it converges to $\frac{Bx^2}{1-x}$. This function is minimal at $x=0$ with value 0 and grows monotonically outwards (we can prove this with calculus if we want to). So setting $x=0.5$ gives 0.5, $x=-0.5$ gives $0.1666\cdots$, so we have a bound of $\frac{B}{2}$ on the interval $[-0.5,0.5]$ which the higher order terms can affect. 
In reality however for most functions the Maclaurin coefficients are not constant, but decline rather fast so we can get a better bound than the geometric series above.
Here is a plot of the shape: Note that the bounding coefficient $B$ is usually quite a bit smaller than 1 in absolute value too, so if we took it into account the blue curve would shrink by the same factor.

A: You can not say that $\sinh(x)=x$. There's a bunch of other terms that do not cancel out, indeed you have
$$\sinh(x)=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots$$
However if $x$ is close to $0$ then $x$ is much larger than $x^3$, $x^5$,... so we can write
$$\sinh(x)=x+x\varphi(x)$$
where $\varphi$ is a function such that $\varphi(x)\rightarrow 0$ when $x\rightarrow 0$. This gives for instance that
$$\lim_{x\rightarrow 0}\dfrac{\sinh(x)}{x}=1$$
Note this a limit as $x\rightarrow 0$ and the Taylor series says nothing interesting about the limit as $x\rightarrow 1$ for instance.
A: If it were to approximate the function $f(x)$, close to some other point, for example around $a$, the Taylor series formula would have been written (in its general form):
$$
f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...
$$
Setting $a=0$ in the above produces:
$$
f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...
$$
which approximates $f$ around $a=0$. It is usually called Maclaurin series and this is the formula which has been used to produce the series of your example (for approximating the functions $f(x)=e^x$ and $f(x)=e^{-x}$ close to $0$). 
