Improper integral convergence

Let 
\[u=x   \text{and}     dv=e^{x}dx.\]
Then 
\[du=dx   \text{and}  v=e^{x}.\]
Using integration by parts 
\[\int xe^{x}dx=\int udv= uv-\int vdu=xe^{x}-\int e^xdx=xe^x-e^{x}+C.\]
Hence
\[\int_{-\infty}^{0} xe^{x}dx=(xe^x-e^x)\big |_{-\infty}^{0}\]
\[=(0-e^{0})- \lim_{x\rightarrow -\infty} (xe^{x}-e^{x})=-1+0=-1.\]
Note that here we are using the fact that 
\[\lim_{x \rightarrow - \infty}xe^{x}=\lim_{x \rightarrow - \infty}\frac{x}{e^{-x}}=\lim_{x \rightarrow - \infty}\frac{1}{-e^{-x}}=\frac{1}{-\infty}=0,\]
where we have used the L'Hopital's Rule. It is also clear that
\[\lim_{x\rightarrow -\infty} e^{x}=e^{-\infty}=\frac{1}{e^{\infty}}=\frac{1}{\infty}=0.\]