Using l'Hôpital rule to find $\lim_{x\to-\infty} xe^x$

Question

I'm trying to solve this limit:

$$\lim_{x\to-\infty} xe^x$$

I'm trying to solve this limit using the l'Hôpital rule. My question is: can I use this rule in the last limit below?

$\lim_{x\to-\infty} xe^x=\lim_{x\to-\infty} \frac{x}{e^{-x}}$

Note the numerator tends to $-\infty$, while the denominator tends to $\infty$, can I use the l'Hôpital rule in this case?

Thanks

Answer 1

Try to compute $\displaystyle \lim_{x \to -\infty}\frac{-x}{e^{-x}}$. Numerator and denominator tend to $\infty$, so you can use l'Hospital rule.

$$\displaystyle \lim_{x \to -\infty}\frac{-x}{e^{-x}}=\lim_{x \to -\infty}\frac{-1}{-e^{-x}}=\lim_{x \to -\infty}\frac{1}{e^{-x}}=0$$, so

$$\lim_{x \to -\infty}\frac{x}{e^{-x}}=-\lim_{x \to -\infty}\frac{-x}{e^{-x}}=0$$

Answer 2

Hint: Your concern is that the top and bottom seem to go to different infinities. Could you multiply in such a way as to get rid of that difference?

Answer 3

You could also try a substitution:

$$y=-x\implies \left(x\to -\infty\implies y\to\infty\right)$$

and your limit becomes

$$\lim_{x\to-\infty} xe^{x}=\lim_{y\to\infty}-ye^{-y}=-\lim_{y\to\infty}\frac y{e^y}$$

and you don't need to worry about signs anymore.