Limit of an integral expression Does anyone have any idea on how to solve this problem ? I don't have the faintest clue on how to even start it . I would sure appreciate a solution , thank you for your patience !

 A: There's probably a better way to do this, but I'll state this for now. First, rewrite
$$
\dfrac{\sin^n x}{\sin^n x + \cos^n x} = \dfrac{1}{1+\cot^n{x}}.
$$
Now, notice that $\cot{x} > 1$ when $0 < x < \pi/4$, $\cot{\pi/4} = 1$, and $0 < \cot{x} < 1$ for $\pi/4 <  x \leq \pi/3$. Therefore, 
$$
\lim_{n\to\infty} \dfrac{1}{1+\cot^n{x}} = \left\{ \begin{array}{cl}
0 & \quad \mbox{if } 0 < x < \pi/4\\
1/2 & \quad \mbox{if } x = \pi/4\\
1 & \quad \mbox{if } \pi/4 \leq x \leq \pi/3
\end{array}\right. 
$$
This is because the $a^n \to \infty$ if $a$ is a fixed number so that $a > 1$ and $a^n \to 0$ if $0 < a < 1$. Let $f(x)$ be this piece-wise function. Thus, by moving the limit inside the integral (which you can do because the integrand is bounded below by 0 and above by 1)
$$
\begin{align*}
\lim_{n\to\infty}\int_0^{\pi/3} \dfrac{\sin^n x}{\sin^n x + \cos^n x} \, dx &= \lim_{n\to\infty} \int_0^{\pi/3} \dfrac{1}{1+\cot^n{x}} \,dx\\
&= \int_0^{\pi/3} \lim_{n\to\infty} \dfrac{1}{1+\cot^n{x}} \, dx\\
&= \int_0^{\pi/3} f(x) \, dx\\
&= \int_0^{\pi/4} f(x) \, dx + \int_{\pi/4}^{\pi/3}f(x) \, dx\\
&= 0 + 1\cdot (\pi/3 - \pi/4)\\
&= \frac{\pi}{12}.
\end{align*}
$$
