In the derivation of the Itô-Doeblin formula, Theorem 4.4.1, we considered only the case of the function $f(x)=\frac{1}{2} x^2$, for which $f^{\prime \prime}(x)=1$. This made it easy to determine the limit of the last term,
$$
\frac{1}{2} \sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2,
$$
appearing in (4.4.5). Indeed,
$$
\begin{aligned}
\lim _{\|\Pi\| \rightarrow 0} \sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2 & =\lim _{\|\Pi\| \rightarrow 0} \sum_{j=0}^{n-1}\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2 \\
& =[W, W](T)=T \\
& =\int_0^T f^{\prime \prime}(W(t)) d t .
\end{aligned}
$$
If we had been working with an arbitrary function $f(x)$, we could not replace $f^{\prime \prime}\left(W\left(t_j\right)\right)$ by 1 in the argument above. It is tempting in this case to just argue that $\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2$ is approximately equal to $\left(t_{j+1}-t_j\right)$, so that
$$
\sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2
$$
is approximately equal to
$$
\sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left(t_{j+1}-t_j\right)
$$