Confused on solving $\lim\limits_{x\to1}{\sqrt{{π\over 2}-\arctan{1\over (x-1)^2}}}$

Question

Calculate $$\lim\limits_{x\to1}{\sqrt{{π\over 2}-\arctan{1\over (x-1)^2}}}$$

Those ones are most of the times tricky limits. Well: $\arctan=\tan^{-1}$, also $\arctan(1)={π \over 4}$. At first, I thought of making them similar fractions by using the denominator $(x-1)^2$.. That option leads to nowhere.

Answer 1

You don't really need to make any simplification at all. Simply recognize that

$$\lim_{t \to \infty} \arctan t = \pi/2$$

and that $1/(x - 1)^2 \to \infty$ as $x \to 1$. Hence, the limit is zero.

Answer 2

Note that by trigonometric identities

$$\lim\limits_{x\to1}{\sqrt{{π\over 2}-\arctan{1\over (x-1)^2}}}=\lim\limits_{x\to1}{\sqrt{\arctan (x-1)^2}}=0$$

Answer 3

$$ x\to 1\implies \frac {1}{(x-1)^2} \to \infty \implies$$ $$ \arctan \frac {1}{(x-1)^2} \to \pi /2 \implies$$

$$\lim\limits_{x\to1}{\sqrt{{π\over 2}-\arctan{1\over (x-1)^2}}}=0$$