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Prove false: $\cot x=o(x^{-1})$ as $x\to 0$

Note: $x_n=o(\alpha_n)$ means $\lim\limits_{n\to\infty}\frac{x_n}{\alpha_n}=0$. Or, for some $\epsilon_n\geq 0$, we have $\epsilon_n\to0$ and $|x_n|\leq\epsilon_n|\alpha_n|$

My attempt: By definition, this is equivalent to $$\lim\limits_{n\to\infty}\frac{\cot x_n}{x_n^{-1}}=\lim\limits_{n\to\infty}\frac{x_n}{\tan x_n}=\lim\limits_{n\to\infty}\frac{1}{\sec^2x_n}=\lim\limits_{n\to\infty}\cos^2x_n$$

What should I do next?

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  • $\begingroup$ So do you know the answer to $\lim_{n \to \infty} \cos^2 x_n$? If it is zero, then your statement is true, else it is false, right? $\endgroup$ – астон вілла олоф мэллбэрг May 29 '18 at 2:57
  • $\begingroup$ @астонвіллаолофмэллбэрг I guess the limit will be 1 because the range of cos^2x is [0,1]? I feel like something's wrong but couldn't tell $\endgroup$ – Thomas May 29 '18 at 3:00
  • $\begingroup$ Answer is correct, reason is wrong : the limit is deduced by using continuity of the function $\cos^2$. Indeed, the answer is then $\cos^2 0 = 1$. $\endgroup$ – астон вілла олоф мэллбэрг May 29 '18 at 3:01

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