lim-x-tan-pix-1-2x-2-x-1-

Question Number 83690 by jagoll last updated on 05/Mar/20

\lim_{x \to \infty} \frac{\tan (\frac{π x + 1}{2 x + 2})}{x + 1} = ?

Commented by mathmax by abdo last updated on 05/Mar/20

l e t f (x) = \frac{t a n (\frac{π x + 1}{2 x + 2})}{x + 1} \Rightarrow f (x) = \frac{t a n (\frac{π}{2} \times \frac{x + \frac{1}{π}}{x + 1})}{x + 1}

= \frac{t a n (\frac{π}{2} \times (\frac{x + 1 + \frac{1}{π} - 1}{x + 1}))}{x + 1} = \frac{t a n (\frac{π}{2} (1 + \frac{1 - π}{π (x + 1)})}{x + 1}

= \frac{t a n (\frac{π}{2} + \frac{(1 - π)}{2 (x + 1)})}{x + 1} = - \frac{1}{t a n (\frac{1 - π}{2 (x + 1)}) (x + 1)}

\Rightarrow f (x) \sim - \frac{1}{\frac{1 - π}{2 (x + 1)} \times (x + 1)} = \frac{- 2}{1 - π} = \frac{2}{π - 1} \Rightarrow

{l i m}_{x \to + \infty} f (x) = \frac{2}{π - 1}

Answered by john santu last updated on 05/Mar/20

\lim_{x \to \infty} \frac{1}{x + 1} \cot (\frac{π}{2} - \frac{π x + 1}{2 x + 2}) =

\lim_{x \to \infty} \frac{\cot (\frac{2 (π - 1)}{4 (x + 1)})}{x + 1} =

\lim_{x \to \infty} \frac{1}{(x + 1) \tan (\frac{π - 1}{2 (x + 1)})} \times \frac{(\frac{π - 1}{2 (x + 1)})}{(\frac{π - 1}{2 (x + 1)})} =

\lim_{x \to \infty} \frac{2 (x + 1)}{(π - 1) (x + 1)} = \frac{2}{π - 1} \Leftarrow the answer

Commented by jagoll last updated on 05/Mar/20

thank you mister

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