The value of $\lim\limits_{x \rightarrow \infty} \frac{x^n+n x^{n-1}+1}{e^{[x]}}, n \in I$ is:

0

$\lim\limits_{x \rightarrow \infty}[x]=\lim\limits_{x \rightarrow \infty} \frac{[x] x}{x}$

$=\lim\limits_{x \rightarrow \infty} \frac{[x]}{x} . \lim\limits_{x \rightarrow \infty} x=\lim\limits_{x \rightarrow \infty} x$  (as $\lim\limits_{x \rightarrow \infty} \frac{[x]}{x}=1$)

$\lim\limits_{x \rightarrow \infty} \frac{x^n+n x^{n-1}+1}{e^{[x]}}$

$=\lim\limits_{x \rightarrow \infty} \frac{x^n+n x^{n-1}+1}{e^x}$

$=\lim\limits_{x \rightarrow \infty} \frac{x^n+n x^{n-1}+1}{1+x+\frac{x^2}{2 !}+...+\frac{x^n}{n !}+\frac{x^{n+1}}{(n+1) !}+...}$

$=\lim\limits_{x \rightarrow \infty} \frac{1+\frac{n}{x}+\frac{1}{x^n}}{\frac{1}{x^n+}+\frac{1}{x^{n-1}}+\frac{1}{2 x^{n-2}}+...+\frac{1}{n !}+\frac{x}{(n+1) !}+\frac{x^2}{(n+2) !}+...}$

$=\frac{1+0+0}{0+0+...+0+\frac{1}{n !}+\infty}=\frac{1}{\infty}=0$

Hence (2) is the correct answer.