+50 While studying the sign of the sequence $(a_n)_n = \left(\frac{1}{n}-\sin \frac{1}{n}\right)^x$ with $x\in\mathbb{R}$ I had the following doubt: since I know that for all $t \geq 0$ it is $t \geq \sin t$ and since $\frac{1}{n} >0$, it is surely $\frac{1}{n} -\sin \frac{1}{n} \geq 0$. However, while I've studied the powers with integer exponent I've never really studied the powers with real exponent. Can someone explain me briefly how can I rigorously deduce (if it is true), from $\frac{1}{n} -\sin \frac{1}{n} \geq 0$, that it is $\left(\frac{1}{n} -\sin \frac{1}{n} \geq 0\right)^x \geq 0$?
Actually I have the same doubts in the other direction: from $\left(\frac{1}{n} -\sin \frac{1}{n} \geq 0\right)^x \geq 0$ can I deduce that $\frac{1}{n} -\sin \frac{1}{n}\geq0$ taking the $x$-th root being careful with the even $x$ in reals as well? Or is this wrong for some reasons?
