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

Question

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?

Answer 1

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

Answer 2

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

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

Answer 3

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

Answer 4

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

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