## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to\pi/6}\sqrt{\csc^2x+5\sqrt3\tan x}=3$$ The function is continuous at $x=\pi/6$.
*Recall Theorem 10: If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$ $$A=\lim_{x\to\pi/6}f(x)=\lim_{x\to\pi/6}\sqrt{\csc^2x+5\sqrt3\tan x}$$ - With function $y=\sqrt x$: For all $x\ge0$, $\lim_{x\to c}\sqrt x=\sqrt c$, so $y=\sqrt x$ is continuous on $[0,\infty)$ That means we can apply Theorem 10 here as long as $\lim_{x\to\pi/6}(\csc^2x+5\sqrt3\tan x)\ge0$. In detail, $$A=\sqrt{\lim_{x\to\pi/6}\csc^2x+5\sqrt3\lim_{x\to\pi/6}\tan x}$$ $$A=\sqrt{\csc^2\frac{\pi}{6}+5\sqrt3\tan\frac{\pi}{6}}=f\Big(\frac{\pi}{6}\Big)$$ $$A=\sqrt{\frac{1}{\sin^2\frac{\pi}{6}}+5\sqrt3\frac{\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}}}$$ $$A=\sqrt{\frac{1}{(\frac{1}{2})^2}+5\sqrt3\frac{\frac{1}{2}}{\frac{\sqrt3}{2}}}$$ $$A=\sqrt{4+5\sqrt3\times\frac{1}{\sqrt3}}=\sqrt{4+5}=\sqrt9=3$$ As shown above, $\lim_{x\to\pi/6}f(x)=f(\pi/6)$, so the function is continuous at $x=\pi/6$