Why is substitution allowed in Taylor Series?

A function $f(x)$ analytic at $x=0$ can be represented as power series within an open disc with radius of convergence $R$. \begin{align*} f(x)=\sum_{n=0}^\infty a_nx^n\qquad\qquad \qquad |x|<R \end{align*}

Any substitution $x=g(u)$ is admissible as long as we respect the radius of convergence. \begin{align*} f(g(u))=\sum_{n=0}^\infty a_n \left(g(u)\right)^n\qquad\qquad\quad |g(u)|<R \end{align*}

We know $f(u)=e^u$ can be represented as Taylor series convergent for all $u\in\mathbb{R}$, i.e. the radius of convergence $R=\infty$. \begin{align*} f(u)=e^u=\sum_{n=0}^\infty \frac{u^n}{n!}\qquad\qquad\qquad u\in \mathbb{R} \end{align*}

Substitution $u=2x$

We consider

\begin{align*} f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad 2x\in \mathbb{R} \end{align*}

This substitution is admissible for all $x \in \mathbb{R}$ since $$2x\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(2x)=e^{2x}$ is $\infty$.

We obtain \begin{align*} f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R} \end{align*}

Substitution $u=x+1$

We consider

\begin{align*} f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x+1\in \mathbb{R} \end{align*}

This substitution is admissible for all $x \in \mathbb{R}$ since $$x+1\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(x+1)=e^{x+1}$ is $\infty$.

We obtain \begin{align*} f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R} \end{align*}

We also obtain \begin{align*} e\cdot e^x&=\left(\sum_{k=0}^\infty \frac{1}{k!}\right)\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty \left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{1}{k!}\cdot\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty \left(\sum_{l=0}^n\frac{1}{(n-l)!}\cdot\frac{x^l}{l!}\right)\\ &=\sum_{n=0}^\infty\left(\sum_{l=0}^n\binom{n}{l}x^l\right)\frac{1}{n!}\\ &=\sum_{n=0}^\infty\frac{(x+1)^n}{n!}\\ &=e^{x+1} \end{align*}

Conclusion: We can use any substitution for convenience as long as we respect the radius of convergence.

If $ f(x)=P_n(x)+x^n\epsilon(x)$ then

$f(u(x))=P(u(x))+(u(x))^n\epsilon(u(x))$

with $\lim_{x\to 0}\epsilon(x)=0$.

thus we need that

$\lim_{x\to 0}\epsilon(u(x))=0$

so, we must have

$$\lim_{x\to 0}u(x)=0$$