Trigonometric limit $0^0$

bring out $(sinx)^n$ from the expression therefore we get

$\lim_{n \to \infty}$ $sinx(1+(cotx)^n)$ $^{\frac{1}{n}}$

= $[(\lim_{n \to \infty}$ $sinx$ ]*[ $\lim_{n \to \infty}$ $(1+(cotx)^n)$ $^{\frac{1}{n}}$ ]

As $cot(\pi/3)$ is less than 1 therefore $\lim_{n \to \infty}$ $(1+(cotx)^n)$ $^{\frac{1}{n}}$ =1

therefore answer is $\boxed{sin(\pi/3)}$ = $\sqrt{3}/2$

$\begin{aligned} L & = \lim_{n \to \infty} \left(\sin^n x + \cos^n x \right)^\frac 1n \\ & = \lim_{n \to \infty} \sin x \left(1 + \cot^n x \right)^\frac 1n & \small \color{#3D99F6} \text{At }x = \frac \pi 3 \\ & = \lim_{n \to \infty} \frac {\sqrt 3}2 \left(1 + \left(\frac 1{\sqrt 3}\right)^n \right)^\frac 1n \\ & = \frac {\sqrt 3}2 \left(1 + 0 \right)^0 \\ & = \boxed{\dfrac {\sqrt 3}2} \end{aligned}$

You can plug in $x=\frac {\pi}{3}$ straight away and from that it is pretty simple. But is plugging in straight away always correct (can it be deceitful) ?