This is from my engineering mathematics textbook. Is this version of taylor's theorem correct ?

Successive Differentiation, Maclaurin's and Taylor's Expansion of Function $-147$

TAYLOR'S THEOREM

Let $f(x)$ be a function of $x$ and $h$ be small. If the function $f(x+h)$ is capable of being expanded in a convergent series of terms of positive integral powers of $h$, then this expansion is given by$$f(x+h)=f(x)+hf^\prime(x)+\dfrac{h^2}{2!}f^{\prime\prime}(x)+\dfrac{h^3}{3!}f^{\prime\prime\prime}(x)+\ldots+\dfrac{x^n}{n!}f^{(n)}(x)+\ldots$$

PROOF:Assume that $$f(x+h)=A_0+A_1h+A_2h^2+A_3h^3+\ldots+A_nh^n+\ldots\tag1$$ where $A$'s are functions of $x$.

Differentiating successively w.r.t $h$, we get $$\begin{align}f^\prime(x+h)&=A_1+2A_2h+3A_3h^2+4A_4h^3+\ldots+ nA_nh^{n-1}+\ldots \\ f^{\prime\prime}(x+h)&=2\cdot A_2+3\cdot 2\cdot A_3 h+4\cdot 3\cdot A_4 h^2+\ldots+n(n-1)A_n h^{n-2}+\ldots\\f^{\prime\prime\prime}(x+h)&=3\cdot2\cdot A_3+4\cdot3\cdot2\cdot A_4 h+\ldots+n(n-1)(n-2)A_nh^{n-3}+\ldots\\\end{align}$$ and, in general, $$f^{(n)}(x+h)=n(n-1)(n-2)\ldots3\cdot2 A_n+\text{ terms in ascending powers of $h$}$$ Putting $h=0$, we get $f^{(n)}(x)=n!\,A_n$ so that $A_n=\dfrac{f^{(n)}(x)}{n!}$

Substituting these value of $A_0,A_1,A_2,\ldots$ in $(1)$, we get $$f(x+h)=f(x)+\dfrac{h}{1!}f^\prime(x)+\dfrac{h^2}{2!}f^{\prime\prime}(x)+\dfrac{h^3}{3!}f^{\prime\prime\prime}(x)+\ldots+\dfrac{h^n}{n!}f^{(n)}(x)+\ldots\tag2$$ Its another form can be obtained by replacing $x$ by $a$ and $h$ by $x-a$, so as to get $$f(x)=f(a)+(x-a)f^\prime(a)+\dfrac{(x-a)^2}{2!}f^{\prime\prime}(a)+\ldots+\dfrac{(x-a)^n}{n!}f^{(n)}(a)+\ldots\tag3$$ The conditions under which the above expansion is valid, are

$ \ $(i) the function $f(x)$ and its derivatives must be finite and continuous in the range of definition of $f(x)$.

$ \ $(ii) the series on the right hand side of $(2)$ must be convergent for which the remainder term $R_n\to0$ as $n\to\infty$

where $R^n=\dfrac{h^n}{n!}f^{(n)}(x+\theta h)$ and $0\lt\theta\lt1$.

In the form $(3)$ of Taylor's expansion, if we take $a=0$ then we have $$f(x)=f(0)+xf^\prime(0)+\dfrac{x^2}{2!}f^{\prime\prime}(0)+\ldots$$ which is nothing but theMaclaurin's expansion of $f(x)$. Thus Macluaurin's expansion is a particular case of Taylor's expansion.

With a slightly different approach we can show here that Taylor's series can be derived from Maclaurin's series.