Solve for the Gammit(gamma +limit)

We know that $\Gamma (\frac{1}{180x})\Gamma (1-\frac{1}{180x})=\dfrac {π}{\sin (\frac{π}{180x})}$ .

Changing $x$ to $h=\frac{1}{x}$ , we get the required limit as

$\dfrac{π}{\frac{π}{180}}\displaystyle \lim_{h\to 0} \dfrac{\frac{πh}{180}}{\sin (\frac{πh}{180})}$

$=180\times 1=\boxed {180}$ .

Similar solution as @Alak Bhattacharya 's just with more details for the better understanding of some.

$\begin{aligned} L & = \lim_{x \to \infty} \frac {\Gamma\left(\frac 1{180x}\right)\Gamma\left(1-\frac 1{180x}\right)}x & \small \blue{\text{By Euler's reflection equation: }\Gamma(z)\Gamma(1-z) = \frac \pi{\sin (\pi z)}} \\ & = \lim_{x \to \infty} \frac \pi{x \sin \frac \pi{180x}} \\ & = \lim_{x \to \infty} \frac {180 \cdot \blue{\frac \pi{180x}}}\blue{\sin \frac \pi{180x}} & \small \blue{\text{Note that }\lim_{u \to 0} \frac {\sin u}u = 1} \\ & = \boxed {180} \end{aligned}$

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Reference: Euler's reflection formula

The first point to notice that we can use the reflection equation .Then we get π/sin(1/x) if we measure it in degrees.Then since limit as x approaches infinity sin(1/x)=π/(180x) .Then inputting it to the π/sin(1/x) yields gives 180x , dividing both side by x gives our answer as 180. (We can show that sin(1/x) is approximately π/(180x), when measured in degrees ,by Maclaurin series)