Comparison Test for Improper Integrals — Question 4

Question 4

Use limit comparison to determine convergence near $x=0$: $$\int_0^1\frac{dx}{\sqrt{x}+x}.$$

Question 4 – Solution

Step 1: Choose a comparison. Near $x=0$, $\sqrt{x}$ is the dominant denominator term, so use $$g(x)=\frac1{\sqrt{x}}=x^{-1/2}.$$ Step 2: Compute the limit ratio. $$\begin{align*} L&=\lim_{x\to0^+} \frac{1/(\sqrt{x}+x)}{1/\sqrt{x}}\\ &=\lim_{x\to0^+}\frac{\sqrt{x}}{\sqrt{x}+x}\\ &=\lim_{x\to0^+}\frac1{1+\sqrt{x}}=1. \end{align*}$$ Step 3: Test the comparison integral. $$\int_0^1x^{-1/2}\,dx=2<\infty.$$ Since $0<L<\infty$, limit comparison shows that the original integral converges. $$\boxed{\text{converges}}$$