How do I prove $\displaystyle\lim_{x\to -∞} (x\cdot e^x)=0$ using L'Hôpital's Rule?

Question

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?

Answer 1

Hint: Try $\frac{x}{1/e^x}$ instead.

Answer 2

Flipping over get \begin{align*} \lim_{x\rightarrow-\infty}xe^{x}&=\lim_{x\rightarrow-\infty}\dfrac{x}{e^{-x}}\\ &=\lim_{x\rightarrow-\infty}\dfrac{1}{-e^{-x}}\\ &=0. \end{align*}

Answer 3

Let $y=-x\to \infty$

$$\lim_{x\rightarrow-\infty}xe^{x}=\lim_{y\rightarrow \infty}-ye^{-y}=\lim_{y\rightarrow \infty}-\frac{y}{e^{y}}\stackrel{H.R.}{=}\lim_{y\rightarrow \infty}-\frac{1}{e^{y}}$$

As an alternative since eventually $e^y\ge y^2$ by squeeze theorem

$$0\le \frac{y}{e^{y}}\le \frac{y}{y^2}=\frac1y \to0$$