How to solve this Limit Algebraically?

Question

$\lim_{x\to\text{-}\infty} xe^x$. This is limit can be easily be seen that it approaches to 0 using graphs. But how to solve it algebraically?

Answer 1

You need to use L'Hospital's Rule. Start by writing $xe^x$ as $\frac{e^x}{1/x}$. See https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

Answer 2

Elementary answer:
(I assume constantly that $x<0$.)
We can reformulate $$ xe^x = x\cdot 2^x \cdot \big(\frac e2\big)^x. $$ We know that $\lim_{x\to\ -\infty}\big(\frac e2\big)^x = 0$, as $\frac e2 > 1$ (we can use it, I suppose?), so it remains to show that $2^x x$ is bounded, which can be done, in fact, by induction.
Obviously, $2^x x < 0$. Also: $$ 2^x x \geq 2^{\lceil x\rceil}\! \lceil x\rceil = -\frac n {2^n} > -1. $$ where I set $n= -\lceil x \rceil$ and the inequality $n < 2^n$ can be proved by induction for every integer $n\geq 0$.
Of course, $2$ isn't in any way significant, anything in $(1,e)$ would do, but the induction proof of the inequality $n<2^n$ is quite standard, I think.

Note: you can also use the inequality $n^2 < e^n$ for every integer $n\geq 0$, also provable by induction, whence $x^2 e^x < 1$ for $x<0$ and, similarly to above: $$ 0 > x e^x = \frac 1x \cdot x^2 e^x > \frac 1x \stackrel{x\to -\infty}\longrightarrow 0. $$