Why are Maclaurin series useful if we can only use them for such a small range of numbers? Okay, I am beginning to get how Maclaurin series work, but what I don't understand is why they are useful. Why would you want an infinite expansion for a series that works for such few values (only between $-1$ and $1$)? It just kind of seems like it isn't very flexible. However, I have a feeling there is some glaring error in my logic that will make me look silly. 
 A: In general a Taylor series tells you what is happening near the point you're expanding around.  To deal with far-away points, you'll want a different expansion, and in many cases that is available.  
In many cases (including rational functions) there will be Laurent series that cover the whole real line except for some isolated points - you will study these if you take complex analysis.
A: A Maclaurin series with a small interval of convergence can be unexpectedly useful if you know a lot about the function. 
The Maclaurin series for $\ln(1+x)$ only converges for $x \in (-1, 1]$, but by using the identity $\ln(xy)=\ln(x)+\ln(y)$, we can actually use it to compute the logarithm of pretty much anything. For example, $3=\left(\frac{3}{2}\right)^2\left(\frac{4}{3}\right)$, and so we can write
$$
\ln 3 = 2 \ln (3/2) + \ln(4/3)
$$
Each term on the right lies in the domain of convergence of the Maclaurin series, so you know how to approximate them. 
One very famous example of this idea is Machin's formula, which applies it to $\tan^{-1}$ instead of $\ln$, and was for centuries the best method anyone knew for approximating $\pi$.
A: Here are some useful scenarios:
1. Computing $\lim_{x\rightarrow 0} \log(1+x)/x$. Expanding the numerator in a series we get the expression $$ \log(1+x)/x = \frac{x - x^2/2 + x^3/3 - ...}{x}$$
$$ = 1 - x/2 + x^2/3 - ...$$
Now taking the limit becomes easy, because all the terms of the infinite sum have an $x$ term except for the first term $1$, so the limit is $1$.


*You may complain that the first one is easily solved using L'Hopital. But try doing that here. We may compute the classical limit $\lim_{n\rightarrow \infty} (1 + 1/n)^n$ 


$$(1 + 1/n)^n = e^{n\log(1 + 1/n)} = e^{n(1/n - (1/n^2)/2 + ...)}$$
$$ = e^{1 - (1/2n) + ..}$$
Now again if you take the limit as $n$ goes large, we get $e$. The radius of convergence for the expansion of $\log(1+x)$ was only $1$ like you said, but it was enough because $1/n$ is eventually smaller than $1$ and the expansion will hold.


*Computing $\log(1/2)$. Write this as $\log(1 + x) = x - x^2/2 + x^3/3 - ...$ and plug in $-1/2$ for $x$. Then evaluating the first three terms we get $\log(1/2) \approx -1/2 - (1/4)/2 -(1/8)/3 = -2/3$. If we wanted more accuracy we could sum more and more terms.  

A: Many calculus problems, for example $d^2y/dx^2=(x+1)y$, just don't have a solution in terms of sine, cos, log, exp and so on.  But the do have solutions as Taylor series.  So we can get a handle on what the solution looks like by plotting the Taylor series.
A: You use the word “useful”. If by this you mean practical application, the answer should be obvious: Nothing, in practice, applies to an actually infinite range of values$\ldots$ even if that that range is very, very large, it is still not infinite. So the “locally” valid Taylor series are just what the doctor ordered. You just have to use enough terms to cover the entire range with the necessary degree of precision. Or, alternately, breaking up the larger interval into a certain number of subintervals, and using the appropriate Taylor expansion for each. Why even use them in the first place ? Because it is so much easier for a computer's CPU, for instance, to evaluate the multiplications and additions involved in calculating a simple polynomial, then to appeal to the unit responsible for transcendental functions, which would be much more time consuming.
