## Thomas' Calculus 13th Edition

$$\lim _{x \rightarrow \pi / 6} \sqrt{\csc ^{2} x+5 \sqrt{3} \tan x}=3$$ The function is continuous at $x=\pi/6$.
Given $$\lim _{x \rightarrow \pi / 6} \sqrt{\csc ^{2} x+5 \sqrt{3} \tan x}$$ \begin{aligned} a) L&=\lim _{x \rightarrow \pi / 6} \sqrt{\csc ^{2} x+5 \sqrt{3} \tan x} \\ &= \sqrt{\csc ^{2} \frac{\pi}{6}+5 \sqrt{3} \tan \frac{\pi}{6} }\\ & =\sqrt{\left(\frac{1}{\sin \frac{\pi}{6}}\right)^{2}+5 \sqrt{3} \cdot \frac{\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}}}\\ &=\sqrt{2^{2}+5 \sqrt{3} \cdot \frac{\frac{1}{2}}{\sqrt{3}}}\\ &=\sqrt{4+5 \sqrt{3} \cdot \frac{1}{\sqrt{3}}}\\ &=\sqrt{4+5}\\ &=3 \end{aligned} Since $$f(x)= \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right)$$ So, we get \begin{aligned}b) f(0)&= \sqrt{\csc ^{2} \frac{\pi}{6}+5 \sqrt{3} \tan \frac{\pi}{6} }\\ & =\sqrt{\left(\frac{1}{\sin \frac{\pi}{6}}\right)^{2}+5 \sqrt{3} \cdot \frac{\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}}}\\ &=\sqrt{2^{2}+5 \sqrt{3} \cdot \frac{\frac{1}{2}}{\sqrt{3}}}\\ &=\sqrt{4+5 \sqrt{3} \cdot \frac{1}{\sqrt{3}}}\\ &=\sqrt{4+5}\\ &=3 \end{aligned} From (a), (b) since $\lim \limits_{x \rightarrow \pi/6} f(t)=f(\pi/6)=3,$ the function is continuous at $x=\pi/6$.