How to find the limit of the following function? I found this task in test $$\lim_{x \to 0+} \frac{1}{x} \int_0^{2x} (\sin{t})^t\,\mathrm{d}t$$. The answer is 2. But I can't find out the algorithm of solving of such an improper integral.

Using the L' Hospital rule we have: $$\lim_{x\to 0_+ } \frac{1}{x}\int_0^{2x} (\sin t)^t dt =\lim_{x\to 0_+ } 2(\sin 2x )^{2x} $$
