Taylor's Theorem with Peano's Form of Remainder
We will prove the result for $h \to 0^{+}$ and the argument for $h \to 0^{-}$ is similar. The proof is taken from my favorite book A Course of Pure Mathematics by G. H. Hardy.
Since $f^{(n)}(a)$ exists it follows that $f^{(n - 1)}(x)$ exists in some neighborhood of $a$ and $f^{(n - 2)}(x)$ is continuous in that neighborhood of $a$. Let $h \geq 0$ and we define another function $$F_{n}(h) = f(a + h) - \left\{f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n - 1}}{(n - 1)!}f^{(n - 1)}(a)\right\}\tag{1}$$ Then $F_{n}(h)$ and its first $(n - 1)$ derivatives vanish at $h = 0$ and $F_{n}^{(n)}(0) = f^{(n)}(a)$. Hence if we write $$G(h) = F_{n}(h) - \frac{h^{n}}{n!}\{f^{(n)}(a) - \epsilon\}\tag{2}$$ where $\epsilon > 0$, then we have $$G(0) = 0, G'(0) = 0, \ldots, G^{(n - 1)}(0) = 0, G^{(n)}(0) = \epsilon > 0\tag{3}$$ Since $G^{(n)}(0) > 0$ it follows that there is a number $\delta_{1} > 0$ such that $G^{(n - 1)}(h) > 0$ for all values of $h$ with $0 < h < \delta_{1}$. Using mean value theorem and noting that $G^{(n - 1)}(0) = 0$ we can see that $G^{(n - 2)}(h) > 0$ for all $h$ with $0 < h < \delta_{1}$. Applying the same argument repeatedly we can see that $G(h) > 0$ for all $h$ wih $0 < h < \delta_{1}$. Thus $$F_{n}(h) > \frac{h^{n}}{n!}\{f^{(n)}(a) - \epsilon\}\tag{4}$$ for $0 < h < \delta_{1}$. Similarly we can prove that $$F_{n}(h) < \frac{h^{n}}{n!}\{f^{(n)}(a) + \epsilon\}\tag{5}$$ for all $h$ with $0 < h < \delta_{2}$.
Thus for every $\epsilon > 0$ there is a $\delta = \min(\delta_{1}, \delta_{2}) > 0$ such that $$\frac{h^{n}}{n!}\{f^{(n)}(a) - \epsilon\} < F_{n}(h) < \frac{h^{n}}{n!}\{f^{(n)}(a) + \epsilon\}\tag{6}$$ for all values of $h$ with $0 < h < \delta$. This proves the theorem for $h \to 0^{+}$.
Slight care has to be taken when dealing with negative values of $h$ for the case $h \to 0^{-}$ because here the nature of inequalities will depend on whether $n$ is odd or even and thus we need to handle both the cases of even $n$ and odd $n$ separately.