What kind of functions cannot be described by the Taylor series? Why is this?
We have the somewhat famous function:
$$f(x)=\begin{cases}e^{-1/x^2}&x\neq 0\\ 0&x=0 \end{cases}$$
is infinitely differentiable at $0$ with $f^{(n)}(0)=0$ for all $n$, so, even though the function is infinitely differentiable, the Taylor series around $0$ does not converge to the value of the function for any $x>0$.
Technically, any function that is infinitely differentiable at $a$ has a Taylor series at $a$. Whether you find that Taylor series useful depends on what you want the series to do.
For example, if given a $g$ infinitely differentiable at $0$, the we know that there exists $C,\epsilon>0$ such that:
$$\left|g(x)-\sum_{k=0}^{n} \frac{g^{(k)}(0)}{k!}x^k\right|<Cx^{n+1}$$
for all $|x|<\epsilon$.
So the finite terms of the Taylor series are in some sense always the "best" polynomial for agreeing with function.
So what happens to our function $f$ above is that $f(x)$ converges to $0$ faster than any function $x^n$.
What we don't always get, for real functions, is a Taylor series that converges to the function in the interval.
In complex numbers, things become intriguing. It turns out, if you define differentiation on complex functions in a relatively simple way, then any function which is differentiable at a point is infinitely differentiable at that point, and the Taylor series converges in some "ball" centered on that point.
If the limit of the Lagrange Error term does not tend to zero (as $n \to \infty $), then the function will not be equal to its Taylor Series.
You can also read more on this in Appendix $1$ in Introduction to Calculus and Analysis $1$ by Courant and John. Hope it helps.
I think the intuition you want is the fact that functions that are not complex-differentiable* (also known as holomorphic) are not described by a Taylor series.
And to give another example that is perhaps even more unexpected than the one given by Andrew:
$$f(z) = \begin{cases} e^{-\frac{1}{z}} && \text{if } z > 0 \\ 0 && \text{otherwise}\end{cases}$$
This function is smooth and zero over an infinitely long interval, and yet nonzero, because it is not holomorphic.
*If you're not familiar with complex differentiation, it's like real differentiation, with $h$ complex:
$$f'(z) = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h}$$
For details, see here.