# Calculate limit of $e^{-2x}/x$ as $x \rightarrow -\infty$

$$\lim _{x\to -\infty }\left(\frac{e^{-2x}}{x}\right)$$

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.

• The short version is exponential growth is way bigger than polynomial, so it diverges. Formally I'd usually show that with L'hospital's (or a power series expansion, which is equivalent). Are you allowed to use the taylor series expansion of $e^x$? – Alan Apr 3 '15 at 9:58

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 $$\lim_{x \to +\infty} \frac{e^x}{x} = +\infty.$$