$g$ is a continuous function. This inequality holds for all $x$ near $1$ except perhaps at $x=1$ itself: $\dfrac{\pi}{2}(x-1)+1\le g(x)\le \tan\left(\dfrac{\pi x}{4}\right)$ **What is the value of $\displaystyle \lim_{x\to 1}g(x)$?** [[☃ input-number 1]]