## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0^+}(\sin x)\sqrt{\frac{1-x}{x}}=0.$$
We will use L'Hopital's rule to calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to0^+}(\sin x)\sqrt{\frac{1-x}{x}}=\lim_{x\to0^+}\frac{\sin x}{\sqrt{\frac{x}{1-x}}}=\left[\frac{\sin 0^+}{\sqrt{\frac{0^+}{1-0^+}}}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0^+}\frac{(\sin x)'}{\left(\sqrt{\frac{x}{1-x}}\right)'}=\lim_{x\to0^+}\frac{\cos x}{\left(\sqrt{\frac{x}{1-x}}\right)'}.$$ Let us calculate the derivative $\left(\sqrt{\frac{x}{1-x}}\right)'$. We will use the chain rule and then the quotient rule: $$\left(\sqrt{\frac{x}{1-x}}\right)'=\frac{1}{2\sqrt{\frac{x}{1-x}}}\left(\frac{x}{1-x}\right)'=\frac{1}{2\sqrt{\frac{x}{1-x}}}\frac{(x)'(1-x)-x(1-x)'}{(1-x)^2}=\frac{1}{2\sqrt{\frac{x}{1-x}}}\frac{1-x+x}{(1-x)^2}=\frac{1}{2\sqrt{\frac{x}{1-x}}}\frac{1}{(1-x)^2}=\frac{1}{2\sqrt{x(1-x)^3}}.$$ Putting this into the limit we have $$\lim_{x\to0^+}(\sin x)\sqrt{\frac{1-x}{x}}=\lim_{x\to0^+}\frac{\cos x}{\frac{1}{2\sqrt{x(1-x)^3}}}=\lim_{x\to0^+}2\cos x\sqrt{x(1-x)^3}=2\cos 0^+\sqrt{0^+(1-0^+)^3}=0.$$