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For example, consider $\lim_{x \to -\infty} \frac{x}{e^x}$, what would constitute as a good justification that the limit diverges too infinity?

It's pretty easy to justify convergent limits rigorously, because one resort to standard results of limits or l'hopital rule, which are all rigorous steps to calculate the limits out. But, unclear to me for divergent.

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    $\begingroup$ The limit $\displaystyle \lim_{x \to \infty} \frac{x^n}{e^x}$ is not a divergent limit. It converges to zero. There are lots of ways to prove this, but the most "mechanical" way is to just apply l'hopital's rule $n$ times. $\endgroup$ Jul 17, 2023 at 11:26
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    $\begingroup$ $\displaystyle \lim_{x \to -\infty} \frac{x}{e^x}$ isn't indeterminant. $\displaystyle \lim_{x \to -\infty} x = -\infty $ and $\displaystyle \lim_{x \to -\infty} \frac{1}{e^x} = \infty $. So the limit of their product is negative infinity. Not sure what more of a "justification" you want? $\endgroup$ Jul 17, 2023 at 12:06
  • $\begingroup$ Would they really be sufficient? Product of infinites is not a very well defined operation to do... @StevenGubkin $\endgroup$
    – Babu
    Jul 17, 2023 at 13:15
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    $\begingroup$ The notation $\displaystyle \lim_{x \to \infty} f(x) = \infty$ actually means that for any $M>0$ you can find an $N>0$ so that if $x > N$ then $f(x) > M$. I am sure you can modify this definition to define the other three cases of $f(x) \to \pm \infty$ as $x \to \pm \infty$. It is easy to prove the "product of infinite limits are infinite" (with appropriate signs) theorems from this definition. If you are teaching an analysis course these proofs are appropriate. If you are teaching an intro calculus course they may or may nor be depending on your school culture. $\endgroup$ Jul 17, 2023 at 14:50
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    $\begingroup$ Linguistic remark: Limits don't converge or diverge. Sequences or functions converge or diverge. (The limit just is a value to which a convergent sequence/function converges.) $\endgroup$ Jul 19, 2023 at 23:21

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You can rewrite $\displaystyle \lim_{x\to -\infty} \frac{x}{e^x}$ as $\displaystyle \lim_{x\to \infty} -x e^x$ by "reversing the x-axis". i.e. making the $x\to -x$ substitution. Then things should be pretty straightforward as both $x$ and $e^x$ to go $\infty$.

The variant that requires more work is $\displaystyle \lim_{x\to +\infty} \frac{x}{e^x}$. I usually go with the heuristic "exponentials grow faster than polynomials" for a non-rigorous explanation for that. (You know, with the appropriate caveats about the base and once you have already talked about $x^n/e^x$ using L'Hopital's rule.)

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