calculus - How do I prove $displaystylelim_{xto -∞} (xcdot e^x)=0$ using L'Hôpital's Rule?

The above limit can be written as: $\displaystyle\lim_{x\to -∞} (x\cdot e^x)=\displaystyle\lim_{x\to -∞} \frac{e^x}{1/x} $.

The limit is an Indeterminate type of ${0/0}$. It can be solved using L'Hôpital's Rule:

$\displaystyle\lim_{x\to -∞} \frac{e^x}{1/x} = \lim_{x\to -∞} \frac{\frac{d}{dx}\left[e^x\right]}{\frac{d}{dx}\left[1/x\right]} = \lim_{x\to -∞} \frac{e^x}{-1/x^2}$

Here the numerator ${e^x\to 0}$ and denominator ${-1/x^2\to 0}$ as ${x\to -∞}$. So after using L'Hôpital's Rule the limit is still an Indeterminate type of ${0/0}$. How do I find a limit that's not an Indeterminate type?