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

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.

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  • $\begingroup$ 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 $e^x$? $\endgroup$ – Alan Apr 3 '15 at 9:58
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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 $$ \lim_{x \to +\infty} \frac{e^x}{x} = +\infty. $$

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