im trying to solve the following by the limit comparison theorem.
Problem: $\int_{0}^{1} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}dx$ is convergent or divergent?
since its a type II with 2 indeterminations I can split them and test them individually. Now, by the limit comparison test its easy to show that $\int_{1/2}^{1} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}\,dx$ is convergent cause we can compare with $\int_{1/2}^{1} \frac{1}{(1-x)^{2/3}}\,dx$ which is convergent and $\lim_{x \to 1} \frac{\tfrac{\sin(x)}{x^{3/2}(1-x)^{2/3}}}{\tfrac{1}{(1-x)^{2/3}}}=\frac{\sin(1)}{x^{3/2}}=1$ then $\int_{1/2}^{1} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}\,dx$ converges.
Now, for the other one, $\int_{0}^{1/2} \frac{\sin(x)}{x^{3/2}(1-x)^{2/3}}\,dx$, I know I can pick $g(x)=\frac{1}{x^{3/2}}$ and it will diverge, but if I take $h(x)=\frac{1}{x^{1/2}}$ it will converge. Which one should I pick and why?