Tessellate S.T.E.M.S (2019) - Mathematics - Category C - Set 3 - Objective Problem 4

Let $g(T) = \int_{0}^{T} \sqrt{1 + (f'(x))^2} dx$ and $h(T) = T$ . The former function is just the arc length of $f(x) = \frac{sin(x)}{x}$ for the entire right half of the curve, which is the length over the entire positive reals as T tends toward infinity. Upon first observation:

$lim_{T \rightarrow \infty} \frac{g(T)}{h(T)} = \frac{\infty}{\infty}$

and applying L'Hopital's Rule yields:

$lim_{T \rightarrow \infty} \frac{g'(T)}{h'(T)} = \sqrt{1 + (f'(T)^2} / 1 = \sqrt{1 + (\frac{T \cdot cos(T) - sin(T)}{T^2})^2} = \sqrt{1 + (\frac{cos(T)}{T} -\frac{sin(T)}{T^2})^2} = \sqrt{1} = \boxed{1}.$