# Show that $G_{1}$ as the space of certain partial sums on $[0,1]$ is equicontinuous

Define $$G_{1}:=\{f_{n}(x):=\sum\limits_{i=1}^{n}\frac{x^{i}}{i!}|n \in \mathbb N\}\subseteq C([0,1])$$

The background to this question is that I want to show that $$G_{1}$$ is relatively compact. It is clear I need to use Arzela-Ascoli but in order to do that I need to prove equicontinuity first, and this is where I am struggling.

I know that $$(f_{n})_{n}$$ converges uniformly on $$[0,1]$$ to some $$f$$ and $$f_{n}(x)\leq \exp(x)$$. From this I know that for any $$\epsilon > 0$$ there exists $$N \in \mathbb N$$ so that $$\vert\vert f_{n}-f\vert\vert_{\infty}<\epsilon$$ for all $$n \geq N$$, where $$\vert\vert f_{n}-f\vert\vert_{\infty}=\sup\limits_{x\in [0,1]}\vert f_{n}(x)-f(x)\vert$$ but how does this help me show continuity let alone equicontinuity.

Let $$\varepsilon>0$$ be given.
Since the limit $$f(x)$$ is a uniformly continuous function on $$[0,1]$$, there exists $$\delta_0>0$$ such that $$|x-y|\le \delta_0 \implies |f(x)-f(y)|\le\varepsilon/3$$.
Since $$f_n\to f$$ uniformly, we can find $$N$$ such that $$\|f-f_n\|_\infty<\varepsilon/3$$ for all $$n\ge N$$. This implies that \begin{align} |f_n(x)-f_n(y)| &\le |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)| \\ &\le \varepsilon/3 + \varepsilon/3 + \varepsilon/3 \\ &= \varepsilon \end{align} whenever $$|x-y|\le \delta_0$$ and $$n\ge N$$.
For the case $$i, each $$f_i$$ is uniformly continuous so we can find $$\delta_i>0$$ such that $$|f_i(x)-f_i(y)|<\varepsilon/3$$. For each such $$i$$ the above calculation checks out whenever $$|x-y|\le \delta_i$$.
Lastly, we take $$\delta = \min\{\delta_0,\delta_1,\dots,\delta_{n-1}\}$$ to get equicontinuity.