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

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?

• 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
• @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
• 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
• @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

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 $$\lim_{x \to \infty} e^{-x}=0.$$

• So, is stating that since it is a product of three functions where limit of one is zero and another two are bounded then it's limit is zero? – Arci Apr 22 '18 at 6:17
• @Anton Yes this is correct. – Olivier Oloa Apr 22 '18 at 6:20
• Thank you so much! – Arci Apr 22 '18 at 6:21
• @Anton You are welcome! – Olivier Oloa Apr 22 '18 at 8:19

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.

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

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

• 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

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

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

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