Do you know exponential

$\begin{aligned} \lim_{x \to \infty} \dfrac{e^x}{P(x)} &= \lim_{x \to \infty} \dfrac{\displaystyle \sum_{r=0}^{\infty} \frac{x^r}{r!}}{\displaystyle \sum_{r=0}^{n} \frac{x^r}{r!}} \\ &= \lim_{x \to \infty} \dfrac{\displaystyle \sum_{r=0}^{n} \frac{x^r}{r!} + \sum_{r=n+1}^{\infty} \frac{x^r}{r!}}{\displaystyle \sum_{r=0}^{n} \frac{x^r}{r!}} \\ &= \lim_{x \to \infty} \left[ 1 + \underbrace{\dfrac{\displaystyle \sum_{r=n+1}^{\infty} \frac{x^r}{r!}}{\displaystyle \sum_{r=0}^{n} \frac{x^r}{r!}}}_{\to \infty \text{ as } x \to \infty} \right] \end{aligned}$

Thus limit diverges and statement is false .

The problem states that $n$ is a large positive integer, however infinity is not a number and so the limit would only equal $1$ if $P(x)$ was an infinite series instead of a partial sum terminating at some large positive integer $n$ . The limit in question is undefined.