$$\lim _{x\to -\infty }\left(\frac{e^{-2x}}{x}\right)$$
The nominator will approach $\infty$ and the denominator will be $-\infty$. I have no idea how to solve this since we end up with a fraction with infinity divided by negative infinity. How do I get started? I should note that I'm not allowed to use lhospital's rule.
There results $$ \lim_{x \to -\infty} \frac{e^{-2x}}{x} = -2\lim_{t \to +\infty} \frac{e^t}{t}. $$ At this stage you must know that the exponential diverges to infinity faster than any power of $t$. It is not really trivial, and in calculus courses we postpone its proof until we can use De l'Hospital's theorem.
If you know that $$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}, $$ then $e^x \geq 1+x+\frac{1}{2}x^2$ for $x>0$, and therefore $$ \lim_{x \to +\infty} \frac{e^x}{x} = +\infty. $$