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How do I show that $\lim\limits_{x \to \infty}x \cos x \neq \infty$ Using the negation of the epsilon delta definition of limit and without using any other theorem?

Meaning that we must find $M>0$ such that for all $N>0$ there exist $x > N$ such that $x \cos x \leq M$.
Thanks.

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Assume that the limit is $\infty$ then for $A>0$ there's $B>0$ such that $$x\cos x>A\quad\text{whenever}\; x>B$$

Let $n\in\Bbb N$ such that $(2n+1)\frac\pi2>B$. Can we find a contradiction with this choice?

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  • $\begingroup$ So close to $50$K!!! ;-) $\endgroup$ – amWhy Mar 29 '14 at 12:05
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take $M=1$ then for all $n\in N$ choose $x=\left(n+1\right)π$ if $n $ is even and $\left(n+2\right)π$ if $n$ is odd then $x\cos \left(x\right)=-x<M=1$ we are done. (by that choices it is clear that $n< x$)

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Assume, towards a contradiction, that for every $K > 0$ there exists $\omega_K$ such that $x > \omega_K \Rightarrow x \cos x > K$. In particular, we can take some $x'>\omega_K$ and we will have $x' \cos x' > K$. Now take $x''=x+\pi$ (assuming radians). Clearly, $x'' > \omega_K$, but is $x'' \cos x'' > K$?

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