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 $x$.