# Why $\lim\limits_{x\to 0} 2x\sin\frac{1}{x}-\cos\frac{1}{x}$ doesn't exist?

I was wondering if there is a rigorous justification for how $$\lim\limits_{x\to 0}2x\sin\frac{1}{x}-\cos\frac{1}{x}$$ does not exist. It's pretty clear that the limit does not exist due to the $$\frac{1}{x}$$ in the trig functions, but I can't really prove that the limit does not exist, since just plugging in $$0$$ to the first term will give an indeterminate form ($$0 \cdot \infty$$). Is there another way to show that the limit does not exist?

The first term is bounded between $$-2|x|$$ and $$2|x|$$, so it approaches zero. For the second, consider the sequences $$a_n = (2n\pi)^{-1}$$, $$b_n = ((2n+1)\pi)^{-1}$$. Along $$a_n$$, the second term is identically $$1$$; along $$b_n$$, it's identically $$-1$$. But since both $$a_n,b_n\to 0$$, the second term has no limit as $$x\to 0$$.
By squeeze law $$]\lim_{x \to 0} 2x \sim(1/x) =0$$ because $$-2x \le 2x \sin(1/x) \le 2x$$ But $$\lim_{x \to 0} \cos(1/x)$$ is real, but uncertain number bounded in $$[-1,1]$$, which does not exists.
You can simplify this $$\lim\limits_{x\to 0} 2x\sin\frac{1}{x}-\cos\frac{1}{x}$$ , using $$y=\frac{1}{x}$$, then $$y\to \infty$$, we can obtain: $$\lim\limits_{y\to \infty} 2\frac{\sin y}{y}-\cos y=\lim\limits_{y\to \infty} 2\frac{\sin y}{y}-\lim\limits_{y\to \infty} \cos y$$, use the fact $$\lim\limits_{y\to \infty} \frac{\sin y}{y}=0$$ because $$\sin$$ is bounded function multiplied by inverse function which is converge to $$0$$, It is well known that $$\cos y$$ dosn't have a limit when $$y\to \infty$$ this means that your limit dosn't exist .