Hi ! Do you know how to solve this problem:
Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point.
I didn't know MathStud28's example. The example that I would have called the "classical" one is: $\displaystyle f(x)=e^{-\frac{1}{|x|}}$ for $\displaystyle x\neq 0$ and $\displaystyle f(0)=0$. I can't remember which famous mathematician introduced it, perhaps Cauchy, anyway it has a long history. And it seems simpler to deal with.
For $\displaystyle x>0$, $\displaystyle f(x)=e^{-\frac{1}{x}}$. First notice that $\displaystyle f(x)\to_{x\to 0^+} 0$. Notice as well $\displaystyle \frac{e^{-\frac{1}{x}}}{x^n}\to_{x\to 0^+}0$ for any integer $\displaystyle n$. To convince yourself of this, substitute $\displaystyle u=\frac{1}{x}$, so that the limit is $\displaystyle \lim_{u\to\infty} u^n e^{-u}$.
Now, the $\displaystyle n$-th derivative of $\displaystyle f$ is (easily) seen to be a rational function of $\displaystyle x$ (a quotient of two polynomials) times $\displaystyle e^{-\frac{1}{x}}$. This is proved by induction (suppose there is a rational function $\displaystyle R$ such that the $\displaystyle n$-th derivative is, for $\displaystyle x>0$, $\displaystyle R(x)e^{-\frac{1}{n}}$, and prove that the next derivative is again of the same kind).
Because of the previously mentioned limit, you obtain that the n-th derivative converges to 0 as $\displaystyle x$ tends to 0 from the right. By symmetry, the same holds from the left.
Finally, there is a theorem (that you probably know if you're asked this problem) telling you that, as a consequence, $\displaystyle f$ is indefinitely differentiable at 0 and the derivatives at 0 are the limits of the derivatives at $\displaystyle x$ when $\displaystyle x$ tends to 0. That is to say, all derivatives at 0 exist and are 0. Yet, the function is only zero at zero...