find-lim-x-0-tan-pi-2-x-x-

Question Number 66321 by mathmax by abdo last updated on 12/Aug/19

f i n d {l i m}_{x \to 0^{+}} {(t a n (\frac{π}{2 + x}))}^{x}

Commented by mathmax by abdo last updated on 24/Aug/19

l e t A (x) = {(t a n (\frac{π}{2 + x}))}^{x} \Rightarrow A (x) = e^{x l n (\frac{π}{2 + x})}

c h a n g e m e n t \frac{π}{2 + x} = t g i v e \frac{2 + x}{π} = \frac{1}{t} \Rightarrow 2 + x = \frac{π}{t} \Rightarrow x = \frac{π}{t} - 2 \Rightarrow

A (x) = B (t) = e^{(\frac{π}{t} - 2) l n (t a n t)} \Rightarrow {l i m}_{x \to 0} A (x) = {l i m}_{t \to \frac{π}{2}} e^{(\frac{π}{t} - 2) l n (t)}

= {l i m}_{t \to \frac{π}{2}} e^{(π - 2 t) \frac{l n (t a n t)}{2 t}} c h a n g e m e n t π - 2 t = u g i v e 2 t = π - u

{l i m}_{t \to \frac{π}{2}} B (t) = {l i m}_{u \to 0} e^{u \frac{l n (t a n (\frac{π}{2} - \frac{u}{2}))}{π - u}}

= {l i m}_{u \to 0} e^{\frac{u}{π - u} l n (\frac{1}{t a n (\frac{u}{2})})} = {l i m}_{u \to 0} e^{- \frac{u}{π - u} l n (t a n (\frac{u}{2}))} = 1

b e c a u s e t a n (\frac{u}{2}) \sim \frac{u}{2} (V (0)) a n d {l i m}_{u \to 0^{+}} u l n u = 0

Commented by mathmax by abdo last updated on 24/Aug/19

a n o t h e r w a y w e h a v e \frac{π}{2 + x} = \frac{π}{2 (1 + \frac{x}{2})} \sim \frac{π}{2} (1 - \frac{x}{2}) = \frac{π}{2} - \frac{π x}{4}

\Rightarrow t a n (\frac{π}{2 + x}) = \frac{1}{t a n (\frac{π x}{4})} b u t {(t a n (\frac{π}{2 + x}))}^{x} = e^{x l n (t a n (\frac{π}{2 + x}))}

\Rightarrow f (x) = e^{- x l n (t a n (\frac{π x}{4}))} \sim e^{- x l n (\frac{π x}{4})} \to 1 (x \to 0)