Comparison Test for Improper Integrals — Question 9

Question 9 – Solution

Step 1: Choose a comparison. Since $\sin x\sim x$ near zero, compare with $g(x)=1/x$.

Step 2: Compute the limit ratio. $$\begin{align*} L&=\lim_{x\to0^+}\frac{1/\sin x}{1/x}\\ &=\lim_{x\to0^+}\frac{x}{\sin x}=1. \end{align*}$$ Step 3: Test the comparison integral. $$\int_0^1\frac{dx}{x} =\lim_{\varepsilon\to0^+}(-\ln\varepsilon)=\infty.$$ Since $0<L<\infty$, limit comparison shows that the original integral diverges. $$\boxed{\text{diverges}}$$