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$\displaystyle \lim\limits_{x \to {\pi/2}} \frac {\sin x -(\sin x)^{\sin x}} {1-\sin x+\log (\sin x)}$

Solution :

We can solve this question by L' Hospital rule

But it will be a bit tedious

Is there any other easy method to solve this question ??

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1  
Yes, Taylor expansion in general. But in this case see below. –  1015 Jun 2 '13 at 15:41
2  
Since this is purely a function of $\sin x$, you can replace it with $$\lim_{y \to 1^-} \frac{y - y^y}{1 - y + \log y}.$$ –  Erick Wong Jun 2 '13 at 15:44
    
@ErickWong thanks –  rst Jun 2 '13 at 15:56

1 Answer 1

up vote 4 down vote accepted

Set $y=\sin x$. Then you have $$\lim_{y\rightarrow 1}\frac{y-y^y}{1-y+\ln y}=\left(\lim_{y\rightarrow 1}y\right) \left(\lim_{y\rightarrow 1}\frac{1-y^{y-1}}{1-y+\ln y}\right)$$

This is now an easier L'Hopital's rule application.

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Very nice observation (+1) –  Babak S. Jun 2 '13 at 15:51
    
Yeah ,it will be easier –  rst Jun 2 '13 at 15:55

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