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

Question Number 203651 by Davidtim last updated on 24/Jan/24

\lim_{x \to 1} {(2 - x)}^{t a n (\frac{π x}{2})} = ?

Answered by Mathspace last updated on 24/Jan/24

c h a n g e m e n t 1 - x = t g i v e

f (x) = {(2 - x)}^{t a n (\frac{π x}{2})}

= f (1 - t) = {(2 - (1 - t))}^{t a n (\frac{π}{2} (1 - t))}

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

= {(1 + t)}^{\frac{1}{t a n (\frac{π t}{2})}} (x \to 1 \Leftrightarrow t \to 0)

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

\sim \frac{2}{π t} \times t (t \in V (0)

= \frac{2}{π} s o

l i m (l n (f (1 - t)) = \frac{2}{π} \Rightarrow

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

\Rightarrow {l i m}_{x \to 1} f (x) = e^{\frac{2}{π}}

Answered by MM42 last updated on 24/Jan/24

1 - x = u

\Rightarrow {l i m}_{u \to 0} {(1 + u)}^{c o t (\frac{π}{2} u)}

\Rightarrow {l i m}_{u \to 0} {({(1 + u)}^{\frac{1}{u}})}^{\frac{u}{s i n \frac{π u}{2}} \times c o s \frac{π u}{2}}

= e^{\frac{2}{π}} ✓

Answered by Calculusboy last updated on 26/Jan/24

S o l u t i o n : b y s u b d i r e c t l y, w e g e t (1^{\infty}) i n d e t e r m i n a n t

l e t y = {(2 - x)}^{t a n (\frac{π x}{2})} (a p p l y l o g t o b o t h s i d e s)

l o g y = l o g {(2 - x)}^{t a n (\frac{π x}{2})}

l o g y = l i \underset{x \to 1}{m t a n} (\frac{π x}{2}) l o g (2 - x)

l o g y = l i \underset{x \to 1}{m} \frac{\frac{d}{d x} [l o g (2 - x)]}{\frac{d}{d x} [\frac{1}{t a n (\frac{π x}{2})}]} \Rightarrow l o g y = l i \underset{x \to 1}{m} \frac{\frac{- 1}{2 - x}}{\frac{d}{d x} [c o t (\frac{π x}{2})]}

l o g y = l i \underset{x \to 1}{m} \frac{\frac{- 1}{2 - x}}{[- \frac{π}{2} {c o s e c}^{2} (\frac{π x}{2})]}

l o g y = \frac{\frac{- 1}{1}}{- \frac{π}{2} l i \underset{x \to 1}{m} {[c o s e c (\frac{π x}{2})]}^{2}}

l o g y = \frac{- 1}{- \frac{π}{2} \cdot 1} \Rightarrow l o g y = \frac{2}{π}

y = e^{\frac{2}{π}}

∴ l i \underset{x \to 1}{m} {(2 - x)}^{t a n (\frac{π x}{2})} = e^{\frac{2}{π}}