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$$\lim_{x \to \infty}\lim_{n \to \infty} {{e^x} \over {x^n}} = ??$$

Now what I thought was by continuing to apply L'Hopitals Rule, till n times, I will get $\lim_{x \to \infty}\lim_{n \to \infty} {{e^x} \over {n!}} $ Now what I think is the numerator is e multiplied infinity times, and the denominator is infinitely large numbers multiplied together many times , so the denominator should dominate and limit $= 0$.

But I have also been taught that $\lim_{x \to \infty}{{e^x} \over {x^n}} = \infty \ \ \forall n \in N$ . So I'm a bit confused. Can someone help me out ??

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Let's limit ourselves to $x>1$. Then $\lim_{n\to\infty}\frac{e^x}{x^n}=0$. As this is true for each $x>1$, we have that $\lim_{x\to\infty}\lim_{n\to\infty}\frac{e^x}{x^n}=0$.

Now this is a good example that you have to be very careful about what you can and cannot do with limits. If you try to swap the limits ($\lim_{n\to\infty}\lim_{x\to\infty}\frac{e^x}{x^n}$), it doesn't work (the inner limit is $\infty$ for every $n$).

There is even a notion of double (simultaneous) limit $\lim_{x\to\infty, n\to\infty}\frac{e^x}{x^n}$, which does not exist (because, if it did, both the above limits of limits would exist and would coincide).

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  • $\begingroup$ Oh ok, so the order also matters for limits !! I really didn't know that, thanks. So does that mean ($\lim_{n\to\infty}\lim_{x\to\infty}\frac{e^x}{x^n}$) does not exist or is it equal to $\infty$ ?? $\endgroup$ – RandomAspirant Sep 10 '19 at 17:52
  • $\begingroup$ @RandomAspirant For infinities we have to introduce conventions how to deal with them. A natural convention here would be to say it is equal to $\infty$, as this is indeed true if the convergence is investigated in a different space (e.g. $\mathbb R\cup\{-\infty,+\infty\}$ - we artifically add two more elements $-\infty$ and $+\infty$ to the set of real numbers). However, it is possible that your book/course presents a different convention, so unfortunately I won't be able to give you a definitive answer here. $\endgroup$ – Stinking Bishop Sep 10 '19 at 18:11
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Hint: $x$ is a constant in $\lim_{n \to \infty} {{e^x} \over {x^n}}$, which you evaluate first.

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