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I'm considering the limit $ \lim_{x\rightarrow -\infty} \frac{x}{e^x}. $ This is an indeterminate form $-\infty/-\infty$ so I should be able to apply l'Hopital's rule to get $ \lim_{x\rightarrow -\infty} \frac{x}{e^x} = \lim_{x\rightarrow -\infty} \frac{1}{e^x}, $ which however is incorrect as the limit should be $-\infty$. What am I missing here?

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    @jamadagni: Yes the precondition has been violated as Thomas has stated in his answer, because $\lim\limits_{x\to - \infty} e^x = 0$, hence the limit is not in indeterminate form, so we can't apply l'Hopital.2012-12-07

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Your limit is not of the $-\infty / -\infty$ form, but of $-\infty / 0$ form. Note that $ \lim_{x\to -\infty} e^x = 0. $ So in all you are taking a very large and negative number and dividing it by a very small (and positive) number. That all gives a very large and negative number. Hence the limit is (as you write) $-\infty$.

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Beside to @Thomas's answer you could use some first terms of the expansion of $\exp(-x)$: $\exp(-x)\cong 1-x+(1/2)x^2-(1/6)x^3+(1/24)x^4-(1/120)x^5$ It is obvious that your limit when $x$ tends to $-\infty$, would be $-\infty$.