Evaluate $\lim_{x \rightarrow \pi/2}\left(\arctan(x/(\tan(x)-x))/(x-\pi/2)\right)?$

Question

I need to evaluate this limit without L'Hospital's Rule but I haven't found a method. $$\lim_{x \rightarrow \pi/2}\left(\arctan(x/(\tan(x)-x))/(x-\pi/2)\right)$$

Answer 1

Note that $\lim_{x\to \pi/2}{\arctan\left(\frac{x}{\tan(x)-x}\right)}=0$. Then this is just the limit definition of the derivative:

$$ \lim_{x\to \pi/2}\frac{\arctan\left(\frac{x}{\tan(x)-x}\right)-0}{x-\pi/2} $$ $$ = \lim_{x\to \pi/2}\frac{d}{dx}{\arctan\left(\frac{x}{\tan(x)-x}\right)} $$A tedious chain-rule calculation produces: $$ = \lim_{x\to \pi/2}\frac{- x \sec^2(x)+\tan(x) }{\tan^2(x)+2 x^2 - 2 x \tan(x)} $$ Multiply through by $\cos^2(x)$ to evaluate the limit. $$ = \lim_{x\to \pi/2}\frac{-x+\sin(x)\cos(x)}{\sin^2(x) +2 x^2\cos^2(x) - 2 x \sin(x)\cos(x)} = \frac{-\pi/2+0}{1+0+0}=\frac{-\pi}{2} $$