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I am having problem in solving this indetermination:

$$\lim_{x\to-\infty} \frac{e^{-x}}{x}$$

I tried to the variable exchange $\frac1x=y , x=\frac1y$ and $x→-∞ , y→+∞ $ but end up with the wrong solution (0)

Could you guys give a hint, please? And also, when it is convenient or not to change the variable? Thank you.

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    You got the wrong solution because if $x\to -\infty$, then $y=\frac1x \to 0^-$. So you will get $$\lim_{y \to 0^-} ye^{-\frac1y}$$ which I don't think it makes it easier. – 2017-02-28
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    The limit does not converge replace $x$ by $-x$: $\lim_{x\to \infty}\frac{e^x}{-x}$, use Bernoulli's limit law (also called the rule of l'Hospital) to show divergence. – 2017-02-28

2 Answers 2

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If $x \to -\infty$, then $-x \to +\infty$ and thus $e^{-x} \to +\infty$. The denominator, which is simply $x$, clearly tends to $-\infty$. So it boils down to the question: which function wins, the exponential or the polynomial?

You can use l'Hôpital's rule on this indeterminate form (*) or perhaps you know that the exponential function dominates any polynomial, i.e. that for any $n \in \mathbb{N}$ you have that: $$\lim_{x \to +\infty} \frac{e^x}{x^n} = +\infty$$ For your limit, this means (you can take $y=-x$ to get this standard limit): $$\lim_{x \to -\infty} \frac{e^{-x}}{x}= -\lim_{y \to +\infty} \frac{e^{y}}{y} = -\infty$$

Alternatively (*), with l'Hôpital's rule: $$\lim_{x \to -\infty} \frac{e^{-x}}{x} = \lim_{x \to -\infty} \frac{-e^{-x}}{1}=-\infty$$

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    Now I see how I have to do the variable exchange. Thank you =) – 2017-02-28
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    Alright, you're welcome! – 2017-02-28
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Apply the change of variables $y = -x$ and rewrite the limit as $$\lim_{x \to -\infty} \frac{e^{-x}}x = -\lim_{y \to +\infty} \frac{e^y}{y} = -\infty$$

You can prove that $$\lim_{y \to +\infty} \frac{e^y}{y} = +\infty$$ by resorting to Bernoulli's inequality.