Suppose we are interested in the interval $[0,\varepsilon]$. An $n$th-order polynomial is a function whose $n$th derivative is constant. So it stands to reason that as long as a function's $n$th derivative doesn't change "too much" over this interval, the function should be well approximated by the first $n$ terms of its Taylor series.
Let's write \[ f^{n}(x) = f^{n}(0) + E_{n}(x), \] where $E_{n}(x)$ is the error function representing how much $f^{(n)}$ differs from constant. Then we can integrate to get \[ f^{(n-1)}(x) = f^{(n-1)}(0) + f^{(n)}(0)x + \int_0^x{E_n(t)dt}, \] where the error in the $(n-1)$th derivative is \[ \left| E_{n-1}(x) \right| = \left| \int_0^x{E(t)dt} \right| \leq \sup_{t \in [0,\varepsilon]}{\left| E_n(t) \right|}x. \] Iterating this process, we eventually find that the original function differs from it's $n$th-order Taylor series by \[ \left| E_0(x) \right| \leq \frac{1}{n!} \sup_{t \in [0,\varepsilon]}{\left| E_n(t) \right|}x^n. \] For the function to be analytic, this expression needs to converge to zero as $n \to \infty$. Due to the $n!$ denominator, this can only fail if $E_n$, and hence the $n$th derivative, is growing extremely quickly as n increases. In this sense, non-analytic functions are pathological.
We can observe this pathological behaviour for our counterexample by plotting a few of the derivatives. While the 3rd derivative (left) shoots up to the value of 30 within .15 of the origin, the 10th derivative (right) is already order $10^{14}$ approximately $.04$ from the origin!
