Find limit $\lim_{x\to\pi/2}\frac{\sin x - (\sin x)^{\sin x}}{1 - \sin x + \ln(\sin x)}$ Find the limit of $$\lim_{x\to\pi/2}\frac{\sin x - (\sin x)^{\sin x}}{1 - \sin x + \ln(\sin x)}.$$
Please help, D L rule does not help
 A: The most elegant solution is what proposed André Nicolas. If you do not do it this way, you would need to apply L'Hospital's rule four times. Taking these derivatives would be a pure nightmare (please, do not try !) and after four derivations, you would see that the limit of the numerator is -6 and the limit of the denominator is -3. Finally, your limit will be 2.  
You told that you do not know Taylor series. Let me briefly explain that to you : it is a way to approximate a function around a given point (something as the tangent at the very first order). When they will teach that topic, give a lot of attention since it is a very powerful tool you will use very often later in your studies. For illustration purposes, I built the expansion of your function : close to x = Pi/2, the function is very close to
2 - 7 (x - Pi/2)^2 / 6
A: Two applications of L'Hospital's Rule will do it. Note that it is enough to find 
$$\lim_{t\to 1} \frac{t-t^t}{1-t+\ln t}.$$
Differentiate top and bottom. On top we get
$$1-(1+\ln t)t^t.\tag{1}$$
At the bottom we get
$$-1+\frac{1}{t}.\tag{2}$$
Multiply both (1) and (2) by $t$, and use L'Hospital's Rule again. 
A: Expanding in Taylor series should help. Use $O$ notation to make it simpler. Do you know how to do these things?
A: Let sinx=t.Thus,limit changes to t->1.then apply DL rule twice to get the answer as 2.
