Question about $e^x$

Question

Let $ p(x)=1+x+x^2/2!+x^3/3!+....+x^n/n!$ where $n$ is a large positive integer.Can it be concluded that $\lim_{x\rightarrow \infty }e^x/p(x)=1$?

Answer 1

No.

$$\frac{e^x}{p(x)}=1+\frac{\sum\limits_{k=n+1}^\infty \frac {x^k}{k!}}{p(x)}\stackrel{x\to\infty}\to\infty$$

Answer 2

Hint: For $x$ large enough, $p(x)<x^{n+1}$. What can we say about $\frac{e^x}{x^{n+1}}$ as $x\to\infty$?

Answer 3

For any polynomial $P$, the limit $$\lim_{x\to\infty}\frac{e^x}{P(x)}$$ is equal to plus or minus $\infty$ (depending on the sign of the leading coefficient of $P$). This statement does not change just because you took the polynomial approximation for $e^x$ as your polynomial $P$.