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Question Number 201221 by Calculusboy last updated on 02/Dec/23

	Answered by MM42 last updated on 02/Dec/23

	$★ ★ ★ {t a n}^{- 1} a - {t a n}^{- 1} b = {t a n}^{- 1} (\frac{a - b}{1 + a b})$ $\Rightarrow {t a n}^{- 1} (\frac{x + 1}{x + 2}) - {t a n}^{- 1} (\frac{x}{x + 2}) = {t a n}^{- 1} (\frac{x + 2}{2 x^{2} + 5 x + 4})$ $\Rightarrow {l i m}_{x \to \infty} {x t a n}^{- 1} (\frac{x + 2}{2 x^{2} + 5 x + 4})$ $= {l i m}_{x \to \infty} \frac{{t a n}^{- 1} (\frac{x + 2}{2 x^{2} + 5 x + 4})}{\frac{1}{x}} \overset{h o p}{=} \frac{1}{2} ✓$
	Commented by Calculusboy last updated on 02/Dec/23

	$n i c e s o l u t i o n s i r$
	Answered by witcher3 last updated on 02/Dec/23
	$t=(1/(x+2))⇒t→0_+ ;x=((1−2t)/t) ⇔lim_(t→0) ((1−2t)/t)[tan^(−1) (1−t)−tan^(−1) (1−2t)] f(z)=tan^(−1) (1−z) over [t,2t]0< t<(1/2)⇒ ∃c∈]t,2t[such That ⇒f(2t)−f(t)=tf′(c)=((−t)/((1−c)^2 +1)) ⇔g(t)=(t/((1−t)^2 +1))≤tan^(−1) (1−t)−tan^(−1) (1−2t)=(t/((1−c)^2 +1))≤(t/(1+(1−2t)^2 ))=h(t) lim_(t→0) ((1−2t)/t)g(t)=(1/2)=lim_(t→0) .((1−2t)/t)h(t)=(1/2) ⇒lim_(t→0) ((1−2t)/t){tan^(−1) (1−t)−tan^(−1) (1−2t)}=(1/2)$
	$t = \frac{1}{x + 2} \Rightarrow t \to 0_{+}; x = \frac{1 - 2 t}{t}$ $\Leftrightarrow \lim_{t \to 0} \frac{1 - 2 t}{t} [\tan^{- 1} (1 - t) - \tan^{- 1} (1 - 2 t)]$ $f (z) = \tan^{- 1} (1 - z) over [t, 2 t] 0 < t < \frac{1}{2} \Rightarrow$ $\exists c \in] t, 2 t [such That$ $\Rightarrow f (2 t) - f (t) = {tf}^{'} (c) = \frac{- t}{{(1 - c)}^{2} + 1}$ $\Leftrightarrow g (t) = \frac{t}{{(1 - t)}^{2} + 1} ⩽ \tan^{- 1} (1 - t) - \tan^{- 1} (1 - 2 t) = \frac{t}{{(1 - c)}^{2} + 1} ⩽ \frac{t}{1 + {(1 - 2 t)}^{2}} = h (t)$ $\lim_{t \to 0} \frac{1 - 2 t}{t} g (t) = \frac{1}{2} = \lim_{t \to 0} . \frac{1 - 2 t}{t} h (t) = \frac{1}{2}$ $\Rightarrow \lim_{t \to 0} \frac{1 - 2 t}{t} {\tan^{- 1} (1 - t) - \tan^{- 1} (1 - 2 t)} = \frac{1}{2}$
	Commented by Calculusboy last updated on 02/Dec/23

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