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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...

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  • $\begingroup$ When you write $p_m\to a$, you mean pointwise convergence ? $\endgroup$ – Gabriel Romon Jul 18 at 12:43
  • $\begingroup$ You are using the symbol $a$ with two distinct meanings. $\endgroup$ – José Carlos Santos Jul 18 at 12:43
  • $\begingroup$ It is true if $x_n(0)=0$ (supposing you have a Maclaurin expansion). $\endgroup$ – Bernard Jul 18 at 12:45
  • $\begingroup$ Sorry, yes, I mean pointwise $\endgroup$ – Alvo Jul 18 at 13:00
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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,$$\lim_{n\to\infty}p_n(x_n)=f(x_0).$$

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    $\begingroup$ Uniform convergence happens for example when each derivative $f^{(n)}$ is bounded by some $M_n$ and $M_n=o((n+1)!)$. $\endgroup$ – Gabriel Romon Jul 18 at 12:57

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