lim-x-x-tan-1-x-1-x-4-pi-4-

Question Number 145806 by bramlexs22 last updated on 08/Jul/21

\lim_{x \to \infty} x (\tan^{- 1} (\frac{x + 1}{x + 4}) - \frac{π}{4}) = ?

Answered by gsk2684 last updated on 08/Jul/21

\lim_{x \to \infty} x (\tan^{- 1} (\frac{\frac{x + 1}{x + 4} - 1}{1 + \frac{x + 1}{x + 4}}))

\lim_{x \to \infty} x (\tan^{- 1} (\frac{- 3}{2 x + 5}))

\lim_{x \to \infty} x (\frac{- 3}{2 x + 5}) = - \frac{3}{2}

Answered by mathmax by abdo last updated on 09/Jul/21

f (x) = x (\arctan (\frac{x + 1}{x + 4}) - \frac{π}{4}) \Rightarrow f (x) =_{x = \frac{1}{t}} \frac{1}{t} (\arctan (\frac{1 + \frac{1}{t}}{4 + \frac{1}{t}}) - \frac{π}{4})

= \frac{1}{t} (\arctan (\frac{t + 1}{4 t + 1}) - \frac{π}{4}) = z (twe have

\arctan (\frac{t + 1}{4 t + 1}) = \arctan (\frac{1}{4} (\frac{4 t + 1 + 3}{4 t + 1}) = \arctan (\frac{1}{4} + \frac{3}{4 (4 t + 1)})

\sim \arctan (\frac{1}{4} + \frac{3}{4} (1 - 4 t)) = \arctan (1 - 3 t)

we have \tan (\arctan (1 - 3 t) - \frac{π}{4}) = \frac{1 - 3 t - 1}{1 + 1 - 3 t} = \frac{- 3 t}{2 - 3 t}

=> z (t) \sim \frac{1}{t} \arctan (\frac{- 3 t}{(2 - 3 t)}) \sim= \frac{1}{t} \frac{- 3 t}{2 - 3 t} \to - \frac{3}{2} (t \to 0) \Rightarrow

\lim_{x \to \infty} f (x) = - \frac{3}{2}