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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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This is not in the form $\infty /\infty $. – glebovg Nov 13 '12 at 17:28
I'm sorry but I am not sure how the other answers have answered the OP's question, which is regarding why L'Hopital's rule doesn't seem to apply in this case. It is true that $$\lim_{x \to -\infty} { x \over {e^x} } = -\infty $$ but if L'Hopital's rule is applied then it would appear that the above limit is equal to $\lim_{x \to -\infty} { 1 \over {e^x} }$ which is however equal to $+\infty$. Has any of the preconditions of L'Hopital's rule been violated here, whereby the calculated limit is faulty? Perhaps is it because the original limit is not indeterminate at all, but in fact it is determi – jamadagni Dec 7 '12 at 3:43
@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. – somebody Dec 7 '12 at 7:20
up vote 7 down vote accepted

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 $-\infty$.

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