Find the value of- $\lim_{x \rightarrow -\infty}\sum_{k=1}^{1000} \frac{x^k}{k!}$ 
QUESTION: Find the value of-  $$\lim_{x \rightarrow -\infty}\sum_{k=1}^{1000} \frac{x^k}{k!}$$


MY ANSWER: Since $x→-\infty$ therefore the summation will look like-
$$ \frac{-\infty^1}{1!}+ \frac{-\infty^2}{2!} + \frac{-\infty^3}{3!} + \frac{-\infty^4}{4!}$$
Now, we know that $${-\infty^{even}}=\infty$$
And $1000$ can be broken down into 500 pairs. And since $k$ is finite the terms will look like $$ -\infty + \infty -\infty +\infty-.....$$
Hence, we can conclude that the overall limit of the summation will tend to zero. 
Am I correct?
Thank you for your help.

One may observe that, for $x\neq0$,
$$
\sum_{k=1}^{1000} \frac{x^k}{k!}=\color{red}{\frac{x^{1000}}{1000!}}\times\left(\color{red}1+\frac{1000!}{x\times999!}+\frac{1000!}{x^2\times998!}+\cdots+ \frac{1000!}{x^{1000}\times1!}\right)
$$ giving that
$$
\begin{align}
\lim_{x \rightarrow -\infty}{\sum_{k=1}^{1000} \frac{x^k}{k!}} &=\lim_{x \rightarrow -\infty} \color{red}{\frac{x^{1000}}{1000!}}\times \lim_{x \rightarrow -\infty}\left(\color{red}1+\frac{1000!}{x\times999!}+\frac{1000!}{x^2\times998!}+\cdots+ \frac{1000!}{x^{1000}\times1!}\right)
\\\\&=\lim_{x \rightarrow -\infty} \color{red}{\frac{x^{1000}}{1000!}}\times \color{red}1
\\\\&=\infty.
\end{align}
$$
