proving $\lim\limits_{x\to \infty} x\cos\frac1x=\infty$ without using arithmetic

Question

proving $$\lim\limits_{x\to \infty} x\cos\frac1x=\infty$$ and $$\lim\limits_{x\to \infty} x\cos x\neq\infty$$ in $\epsilon,\delta$ form without using arithmetic

i am trying to prove that for every $M>0$ there is $N>0$ that for every $X$ that sustains $X>N$ so $f(x)>M$

and for $\lim\limits_{x\to \infty} x\cos x\neq\infty$ i wanna prove that $f(x)<M$

i am trying to start from

$x\cos\frac1x>N$

and i really dont know were to take it

Answer 1

For the first one observe that if $x>1$ then $0<\frac{1}{x}<1$ and since $\cos x$ is monotonic increasing on $[0,1]$ you have that if $x>1$ that $x\cos \frac{1}{x} > x \cos{1} > \frac{x}{2}$ so for any $M>\frac{1}{2}$ choose $N = 2M$ and for all $x>N$ you have that $x\cos \frac{1}{x} > M$

For the second one observe that if you add $\pi$ to x than you get $-\cos(x)$ since it is alternating the limit cannot exist. In fact there are cluster points on the entire real line for the function.

Answer 2

As $x \rightarrow \infty$, $\frac{1}{x} \rightarrow 0$. So by continuity of $\cos$ at $0$, $ \quad \cos(1/x) \rightarrow \cos(0) = 1$. On the other hand, obviously as $x \rightarrow \infty$ we have that $x \rightarrow \infty$ !.

So it remains for you to show the simple result that if

I $\quad \lim_{x \rightarrow \infty} f(x) = 1$ and

II $\quad \lim_{x \rightarrow \infty} g(x) = \infty$

then $\quad \lim_{x \rightarrow \infty} f(x)g(x) = \infty.$

Proof of this result: let $N \in \mathbb{N}$. By I $\exists K \in \mathbb{R}$ such that for all $x \geq K, f(x) \geq 1/2$. By II $\exists K' \in \mathbb{R}$ such that for all $x \geq K', g(x) > 2N$. Put $K^{\ast} := \max\{K,K'\}$. Then for all $x \geq K^{\ast}$ we have $f(x)g(x) > 2N/2 = N$. Since $N$ was an arbitrary natural number we are done.

Answer 3

First one : You begin by stating that for $0\leq x\leq \pi/2$, you have $\cos x\geq1-2x/\pi$. So for $2/\pi\leq x\leq +\infty$, $cos(1/x) \geq 1-\frac{2}{x\pi}$. Multiplying by x, you get $x\cos \frac{1}{x}\geq x-\frac{2}{\pi}$ (for each $x\geq\frac{2}{\pi}$, so in particular, for any x>M).

For the second one, you can use $u_n = 2n\pi$ and $v_n = 2n\pi+\pi/2$. Calculating the general term, you exhibit 2 series with a $+\infty$ limit, yet $f(u_n) \neq f(v_n)$. Proving that there is no limit.