Continuity of $g(\theta) = \frac{1}{2\pi^2\theta^3}-\frac{\pi}{2}\cot(\pi\theta)\csc^2(\pi\theta)$ at $\theta=0$ I have the following function which I'm considering on $[0,1)$
$$g(\theta) = \frac{1}{2\pi^2\theta^3}-\frac{\pi}{2}\cot(\pi\theta)\csc^2(\pi\theta).$$
According to a graph in mathematica it is continuous at $$\theta=0.$$I want to prove this rigorously by showing it has a finite limit. I've thought about using L'Hopital's rule but the expressions are too complicated. Any suggestions? 
 A: Hint We have
$$\cot(\pi t)=\frac{1}{\pi t}-\frac{1}{3}\pi t+O(t^3)$$
$$\csc^2(\pi t)=\frac{1}{\pi^2 t^2}+\frac{1}{3}+O(t^2)$$
so we find
$$g(t)=_{t\to 0}O(t)$$
and thus 
$$\lim_{t\to 0}g(t)=0.$$
A: Note that $g(\theta)=\pi h(\pi\theta)/k(\pi\theta)$ with $k(x)=2x^3\sin^3(x)$ and $h(x)=\sin^3(x)-x^3\cos(x)$. 
First, one term in the expansion of sine at zero is $\sin(x)=x+O(x^3)$, which yields $k(x)\sim2x^6$. Second, two terms in the expansion of cosine at zero are $\cos(x)=1-x^2/2+O(x^4)$. Third, two terms in the expansion of sine at zero are $\sin(x)=x-x^3/6+O(x^5)$, which yield 
$$
\sin^3(x)=(x-x^3/6+O(x^5))^3=x^3(1-x^2/6+O(x^4))^3=x^3(1-x^2/2+O(x^4)).
$$
Thus, $h(x)=x^3\cdot O(x^4)=O(x^7)$, hence $h(x)/k(x)=O(x)$. This implies in particular that $g(\theta)\to0$ when $\theta\to0$.
Edit: And $g'(0)=\frac\pi{30}\ne0$ hence $g(\theta)=\Theta(\theta)$.
A: Or; when $\theta$ tends to $0$, we know that $$\tan(\pi\theta)\sim\pi\theta,~~ \sin(\pi\theta)\sim\pi\theta$$ Hence your function tends to zero.
