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Function: $e^{-2x}x\cos(x^2)$ if x tends to +infinity. I've tried to calculate it but I got 0*infinity, it's indeterminate.

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    $\begingroup$ Do you mean $$\lim_{x \to \infty}e^{-2x}xcos(x^2)=?$$ $\endgroup$ – Khosrotash Sep 27 '16 at 18:33
  • $\begingroup$ hint $\cos(x)\leq 1$ $\endgroup$ – tired Sep 27 '16 at 18:36
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Since $\cos(x^2)$ is bounded and $\lim_{x\to\infty}xe^{-2x}=0$:

$$\lim_{x\to\infty}x\cos(x^2)e^{-2x}=0$$

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Since $-1\leq\cos(x^2)\leq 1$ for all $x$ we have $$0 = \lim_{x\rightarrow\infty}-xe^{-2x}\leq \lim_{x\rightarrow\infty}\cos(x^2)x e^{-2x}\leq \lim_{x\rightarrow\infty}xe^{-2x} = 0$$ so $\lim_{x\rightarrow\infty}\cos(x^2)x e^{-2x} = 0$ by the squeeze theorem

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