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I was studying analysis and came across this limit. I already had a practice in calculus, so I am familiar with basic methods of computing the limits; but as I understand here I was expected to do more "rigorous" and precise job in finding it. Like using the definition of limit or analyzing the expression itself. Since it looks a bit complicated, I wanted to know what would be the best way of computing it:

$$\lim\limits_{x\to +\infty} \frac{x^{2}e^{-x}}{x^{2}+1}\sin(xe^{x^{2}})$$

Also, is it possible(and practical) to find it by using the $\varepsilon-\delta$ definition?

Arci
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    Just for the records: since the $\epsilon$-$\delta$-definition is the definition of limits, every limit can be computed using it. It just can get arbitrarily complex and this is why we developed techniques to find the limit easiert. – M. Winter Apr 22 '18 at 16:47
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    @M.Winter: I disagree. The term "compute" does not mean "prove". In fact, there are some methods to prove limits by the ε-δ definition that require knowing the limit value, without which the method doesn't work. – user21820 Apr 22 '18 at 17:11
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    Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Apr 22 '18 at 17:17
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    @user21820 You are right. "Prove" would be the better term. On the other hand, all other techniques are just educated guesses about the limit which then is checked with the $\epsilon$-$\delta$-definition. – M. Winter Apr 22 '18 at 17:30
  • @MartinSleziak, I apologize for vagueness and shortness of the question; fixed it and added a little bit of details. Hope it fits the standards now. – Arci Apr 22 '18 at 18:45
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    @Anton I highly appreciate your willingness to modify your answer. I will upvote your answer. Always remember to add context before posting your question. +1. – Sarvesh Ravichandran Iyer Apr 23 '18 at 04:16
  • @M.Winter: You're wrong about that. There are very general algorithms for symbolic limit computation. Let me emphasize, "compute" does not mean "prove". Whether some computation algorithm gives a correct symbolic representation of the limit value requires proof, but using the algorithm does not at all involve proving. See this post and the linked ones for a sketch of such an algorithm that is actually used by real computer algebra systems since quite a long time ago. – user21820 Apr 23 '18 at 05:03
  • @M.Winter: So, no, these are not educated guesses. And, no, they are not checked by ε-δ definition, but by chaining theorems. The most basic theorems may use the ε-δ definition, but nobody is so silly as to resort to the ε-δ definition for other theorems. – user21820 Apr 23 '18 at 05:04
  • Since the question has not been reopened, here is my answer to your question: is it possible(and practical) to find it by using the ε−δ definition? No, it is impractical, and this would be obvious from my above comment. A weak analogy is that attempting to find a limit value such that the ε-δ definition can be proven is like attempting to build a skyscraper by using bare hands to lay the bricks or concrete blocks one at a time. It 'can' be done, but it is far easier to use your hands to invent machines to help you. Limit theorems are like those machines. – user21820 Apr 24 '18 at 10:35
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    @user21820 : you have already expressed all that I would have said here. The analogy with machines will probably help many readers. – Paramanand Singh Apr 28 '18 at 08:03

4 Answers4

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Hint. One has $$ \left|\frac{x^{2}}{x^{2}+1}\right|\le1, \qquad \left|\sin\left(xe^{x^{2}}\right)\right|\le1,\qquad x \in \mathbb{R}, $$ and $$ \lim_{x \to \infty} e^{-x}=0. $$

Mutantoe
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Olivier Oloa
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As other answers explained, the key idea is that $\left|\frac{x^2e^{-x}}{x^2+1}\sin(xe^{x^2})\right|\le e^{-x}$ from which it easily follows, using the squeeze theorem, that the limit is 0.

I'd like to also address your question about the ε−δ definition.

It is always "possible" to prove limits directly using the ε−δ definition without any theorems. That's the nature of mathematical proofs - you can simply unroll the proof of whichever theorem you were using. (I'm saying "prove" rather than "compute" because the definition allows you to discern whether something is the limit or not, it doesn't give you tools to figure out what is the limit in the first place).

But such direct proofs are often extremely cumbersome and convoluted - sometimes so much so that it's impractical to put them on paper. The theorems save a lot of work.

However, this is not such a case - since most of the function is fluff, a direct proof is manageable.

We will prove that the limit is $L=0$. Let $\epsilon>0$. Take $M=-\log(\epsilon)$ ($M$ is the analogue to $\delta$ in limits where $x\to\infty$). For every $x>M$, you have

$\left|\frac{x^2e^{-x}}{x^2+1}\sin(xe^{x^2})-L\right| = \left|\frac{x^2e^{-x}}{x^2+1}\sin(xe^{x^2})\right| \le e^{-x} < e^{-M}=e^{\log\epsilon}=\epsilon$

Meni Rosenfeld
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    This is the only answer that remotely tries to address the actual question. The OP never even asked how to find the limit, they asked how could they find limits like this (potentially even knowing how to compute this specific limit). – Git Gud Apr 22 '18 at 11:15
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We have that

$$ -\left|\frac{x^{2}e^{-x}}{x^{2}+1}\sin(xe^{x^{2}})\right|\le \frac{x^{2}e^{-x}}{x^{2}+1}\sin(xe^{x^{2}})\le \left|\frac{x^{2}e^{-x}}{x^{2}+1}\sin(xe^{x^{2}})\right|$$

and

$$0 \le \left|\frac{x^{2}e^{-x}}{x^{2}+1}\sin(xe^{x^{2}})\right|\le \frac{x^{2}}{x^{2}+1}\cdot e^{-x}\to 1 \cdot 0=0$$ therefore by squeeze theorem

$$\frac{x^{2}e^{-x}}{x^{2}+1}\sin(xe^{x^{2}})\to 0$$

user
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Hint: $$-\frac{1}{e^x}<\frac{x^2}{x^2+1}\cdot \frac{1}{e^x}\cdot \sin{\left(xe^{x^2}\right)}<\frac{1}{e^x}.$$

farruhota
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