lim-x-pi-2-1-sin-x-tan-x-2-1-

Question Number 162301 by mnjuly1970 last updated on 28/Dec/21

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

Answered by Mathspace last updated on 28/Dec/21

f (x) = {(1 - s i n x)}^{t a n (\frac{x}{2}) - 1}

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

f (x) = f (t + \frac{π}{2})

= (1 - s i n {(t + \frac{π}{2})}^{t a n (\frac{t}{2} + \frac{π}{4})}

= {(1 - c o s t)}^{t a n (\frac{t}{2} + \frac{π}{4})}

= e^{t a n (\frac{t}{2} + \frac{π}{4}) l n (1 - c o s t)} (t \to 0)

= e^{\frac{t a n (\frac{t}{2}) + 1}{1 - t a n (\frac{t}{2})} l n (1 - c o s t)}

c o s t \sim 1 - \frac{t^{2}}{2} \Rightarrow 1 - c o s t \sim \frac{t^{2}}{2}

\frac{t a n (\frac{t}{2}) + 1}{1 - \tan (\frac{t}{2})} \sim \frac{\frac{t}{2} + 1}{1 - \frac{t}{2}} \sim 1 \Rightarrow

f (t + \frac{π}{2}) \sim e^{l n (\frac{t^{2}}{2})} = e^{2 l n t - l n 2}

\to 0 \Rightarrow {l i m}_{x \to \frac{π}{2}} f (x) = 0