lim-x-pi-2-2-cos-x-1-x-x-pi-2-

Question Number 131769 by liberty last updated on 08/Feb/21

\lim_{x \to π / 2} \frac{2^{- \cos x} - 1}{x (x - \frac{π}{2})} = ?

Answered by bemath last updated on 08/Feb/21

\lim_{x \to π / 2} \frac{1}{x} . \lim_{x \to π / 2} \frac{2^{\sin (x - \frac{π}{2})} - 1}{x - \frac{π}{2}} =

\frac{2}{π} . \lim_{h \to 0} \frac{2^{\sin h} - 1}{h} = \frac{2}{π} . \lim_{h \to 0} \frac{\ln 2 . 2^{\sin h} . \cos h}{1}

= \frac{2 \ln 2}{π} = π^{- 1} . \ln 4

Answered by Dwaipayan Shikari last updated on 08/Feb/21

\lim_{x \to \frac{π}{2}} \frac{2^{- c o s (x)} - 1}{x (x - \frac{π}{2})} = \frac{2^{- s i n (\frac{π}{2} - x)} - 1}{x (x - \frac{π}{2})} = \frac{1 - (\frac{π}{2} - x) l o g (2) - 1}{\frac{π}{2} (x - \frac{π}{2})}

= \frac{2 l o g (2)}{π} = \frac{l o g (4)}{π} \lim_{x \to 0} a^{x} = 1 + x l o g (a)

Answered by mathmax by abdo last updated on 08/Feb/21

f (x) = \frac{2^{- cosx} - 1}{x (x - \frac{π}{2})} we do the changement x - \frac{π}{2} = t (so t \to o) and

f (x) = f (t + \frac{π}{2}) = \frac{2^{sint} - 1}{t (t + \frac{π}{2})} we have 2^{sint} = e^{sint \ln (2)}

sint \sim t - \frac{t^{3}}{6} \Rightarrow e^{\ln (2) sint} \sim e^{\ln (2) (t - \frac{t^{3}}{6})} \sim 1 + \ln (2) (t - \frac{t^{3}}{6}) \Rightarrow

f (t + \frac{π}{2}) \sim \frac{\ln (2) (t - \frac{t^{3}}{6})}{t (t + \frac{π}{2})} = \frac{\ln (2) (1 - \frac{t^{2}}{6})}{t + \frac{π}{2}} \to \frac{2 \ln (2)}{π} (t \to 0) \Rightarrow

\lim_{x \to \frac{π}{2}} f (x) = \frac{2 \ln (2)}{π}