Let's say $p_n (z)$ is the Taylor-expansion of a function $a(z)$ up to the $n$-th order. (Consider $a:\mathbb{R} \rightarrow \mathbb{R}$). Now I have a series $x_n\rightarrow x$ for $n \rightarrow\infty.$ I know that $p_m \rightarrow a$ and $x_n \rightarrow x $ but is it true that $$p_n (x_n) \rightarrow a(x)$$ ?
On the one hand this seems totally clear to me but I can't come up with a proper argument...
If the sequence $(p_n)_{n\in\mathbb N}$ converges uniformly to $a$ in an interval $[\alpha,\beta]$, if $(\forall n\in\mathbb N):x_n\in[\alpha,\beta]$, and if $\lim_{n\to\infty}x_n=x_0$, then, yes,$$\lim_{n\to\infty}p_n(x_n)=f(x_0).$$