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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?

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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 $x\to 0$.

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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.

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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 .

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