$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$

Question

$$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$$

I tried this

$\frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1} = 1-2\sqrt{\frac{\arctan(x)-\frac{\pi}{4}}{x-1}}.\frac{1}{\sqrt{x-1}}$

I am stuck here !

Answer 1

Let's compute $$ \lim_{x\to1^+}\frac{\sqrt{\arctan x-\frac{\pi}{4}}}{x-1} $$

One way could be to substitute $$ \sqrt{\arctan x-\frac{\pi}{4}}=t $$ so $$ \arctan x=t^2+\frac{\pi}{4} $$ and therefore $$ x=\tan(t^2+\tfrac{\pi}{4})=\frac{\tan(t^2)+1}{1-\tan(t^2)}, \qquad x-1=\frac{2\tan(t^2)}{1-\tan(t^2)} $$ so we get $$ \lim_{t\to0^+}\frac{t(1-\tan(t^2))}{2\tan(t^2)}= \lim_{t\to0^+}\frac{1}{2t}\frac{t^2}{\tan(t^2)}(1-\tan(t^2)) $$ The first factor goes to $\infty$, the other two go to $1$, so the overall limit is $\infty$.

Now $$ \lim_{x \to 1^+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}= \lim_{x\to1^+}\biggl(1-\frac{\sqrt{\arctan x-\frac{\pi}{4}}}{x-1}\,\biggr) =-\infty $$

A different way is considering that $$ \lim_{x\to1^+}\frac{\arctan x-\frac{\pi}{4}}{x-1} $$ is the derivative of $\arctan$ at $1$, so it is finite (precisely $1/2$); then our limit is $$ \lim_{x\to 1^+}\frac{\arctan x-\frac{\pi}{4}}{x-1} \frac{1}{\sqrt{\arctan x-\frac{\pi}{4}}}=\infty $$

Answer 2

$$ \begin{aligned} \lim_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1} & = \lim _{t\to 0^+}\left(\frac{t+1-\sqrt{\arctan \left(t+1\right)-\frac{\pi }{4}}-1}{t+1-1}\right) \\& = \lim _{t\to 0^+}\left(\frac{t-\sqrt{\arctan \left(t+1\right)-\frac{\pi \:}{4}}}{t}\right) \\& = \lim _{t\to 0^+}\left(\frac{2t-\sqrt{4\arctan \left(t+1\right)-\pi \:}}{2t}\right) \\& = \lim _{t\to 0^+}\left(\frac{2t-\sqrt{\pi +2t-t^2+o\left(t^2\right)-\pi \:\:}}{2t}\right) \\& = \lim _{t\to 0^+}\left(\frac{2t-\sqrt{-t^2+2t}}{2t}\right) \\& =\frac{1}{2} \lim _{t\to 0+}\left(2-\sqrt{\frac{2}{t}-1}\right) \\& = \color{red}{-\infty} \end{aligned} $$

Solved with substitution $t = x-1$ and Taylor expansion