As $1/x$ is continuous, you need to calculate $\lim_{M\to \infty} \sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}.$ We have $ \frac{1}{M^i} \geq \frac{M!}{(M+i)!} $ independent of $x$. Thus, $\frac{M}{M-x}=\sum_{i=0}^\infty \frac{1}{M^i} x^i \geq\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i} .$ With $M\to\infty$, we find that the limit of the sum is smaller or equal to 1 for all $x$.
To find a lower bound, we just take the term corresponding to $i=0$ (all terms are positive), and we have $\sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i} \geq 1.$
Concluding, we have that $\lim_{M\to \infty} \sum_{i=0}^{\infty}\frac{M!}{\left(M+i\right)!}x^{i}=1$ so your limit is also 1 independent of $x$.