I have the limits and I am wondering why it convergs?
$$\lim_{x \rightarrow \infty} e^{-x} \Big(x^3 -3x \Big)$$ also it converges to $0$ although it equals $(0 \times \infty) \neq 0$
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@Arthur really I was not aware! so we can keep applying l'hopital until we get rid of the fraction $\infty / \infty$. I didn' exactly understand the second part! what can I do more? – user0007 Nov 18 '19 at 16:34
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In THIS ANSWER, I showed that for all $n\in \mathbb{N}$, $n\ge1$, and $x>-1$,
$$e^x>\left(1+\frac xn\right)^n \tag1$$
Setting $n=4$ in $(1)$, it is evident that $e^x>\frac{x^4}{256}$. Thus, we have for $x>1$
$$\begin{align} 0\le e^{-x}(x^3-x)&=\frac{x^3-x}{e^x}\\\\ &<\frac{x^3-x}{\frac{x^4}{256}}\\\\ &=\frac{1}{256}\left(\frac{1}{x}-\frac{1}{x^3}\right)\tag2 \end{align}$$
Letting $x\to\infty$ in $(2)$ and applying the squeeze theorem, we find that
$$\lim_{x\to\infty}e^{-x}(x^3-x)=0$$
Mark Viola
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By squeeze theorem for the first one
$$\frac{2x}{1+x^2}\le \frac{2x}{x^2}=\frac 2 x \to 0$$
and for the second one since eventually as $x\to \infty$ we have $e^x \ge x^4$
$$e^{-x} \Big(x^3 -3x \Big)=\frac{x^3 -3x}{e^x} \le \frac{x^3 -3x}{x^4}\to 0$$
user
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