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?