# Showing that $\lim\limits_{x \to \infty}x \cos x \neq \infty$.

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.

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?
• So close to $50$K!!! ;-) – amWhy Mar 29 '14 at 12:05
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 $$x\cos \left(x\right)=-x we are done. (by that choices it is clear that $$n< x$$)
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 $x'' \cos x'' > K$?