Your answer is correct provided that the limit $\lim_{x^-\to 2}\sum_{i=1}^n(f_x(i)\cdot x)^i$ is uniform with respect to n. The reason for this is the following theorem. And their justification consists in applying the triangle inequality exhaustive.
Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : X \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $X$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $X$ over the base $\mathcal{B}_{T}$ to a function $F : X \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(x)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))$ exist and the equality
$ \lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))=\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x)) $ holds.
This theorem can be found in books of Lang (Analysis vol. 2 for example). But I think the wording of Zorich (Mathematical Analysis II p. 381) more generally.
The converse of this theorem is also true.
In this case we have: $T=\mathbb{N}\cup\{\infty\}$, $X=[1,2]$, $\mathcal{B}_{T}=\mathcal{B}_{\mathbb{N}\cup\{\infty\}}$ the set of parts of $\mathbb{N}\cup\{\infty\}$.
and $F_n(x)=\Sigma_{i=1}^n( f(i)\cdot x)^i$. If we prove that $ |F_n(y)-F_n(x)|0 $ we get what we wanted. I think using the mean value theorem for $ F_n (x) $ and using appropriate cotes to F '(x) with $ 2 - \epsilon we can prove that $ L = 1$.
Or if $\lim_{x\to 2}F_n(x)$ is dependent of n so justification is not worth! And the limit does not exist.