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

$$$$$$\lim {}_{x\to \frac{\pi }{2}}{(1-sin\left(x\right))}^{(tan\left(\frac{x}{2}\right)-1)}=?$$$$$$

Answered by Mathspace last updated on 28/Dec/21

$$f\left(x\right)={(1-sinx)}^{tan\left(\frac{x}{2}\right)-1}$$$$changementt=x-\frac{\pi }{2}give$$$$f\left(x\right)=f(t+\frac{\pi }{2})$$$$=(1-sin{(t+\frac{\pi }{2})}^{tan(\frac{t}{2}+\frac{\pi }{4})}$$$$={(1-cost)}^{tan(\frac{t}{2}+\frac{\pi }{4})}$$$$=e^{tan(\frac{t}{2}+\frac{\pi }{4})ln(1-cost)}(t\to 0)$$$$=e^{\frac{tan\left(\frac{t}{2}\right)+1}{1-tan\left(\frac{t}{2}\right)}ln(1-cost)}$$$$cost\sim 1-\frac{t^2}{2}\Rightarrow 1-cost\sim \frac{t^2}{2}$$$$\frac{tan\left(\frac{t}{2}\right)+1}{1-\tan \left(\frac{\mathrm{t}}{2}\right)}\sim \frac{\frac{t}{2}+1}{1-\frac{t}{2}}\sim 1\Rightarrow$$$$f(t+\frac{\pi }{2})\sim e^{ln\left(\frac{t^2}{2}\right)}=e^{2lnt-ln2}$$$$\to 0\Rightarrow {lim}_{x\to \frac{\pi }{2}}f\left(x\right)=0$$

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