Answer
\begin{aligned} \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right) =\frac{\sqrt{2}}{2} \end{aligned}
The function is continuous at $ t=0$.
Work Step by Step
Given $$ \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right)$$
\begin{aligned} \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right) &=\cos \left(\frac{\pi}{\sqrt{19-3 \sec 0}}\right)\\ &=\cos \left(\frac{\pi}{\sqrt{19-3}}\right)\\ &=\cos \left(\frac{\pi}{4}\right)\\ &=\frac{\sqrt{2}}{2} \end{aligned}
Since $$ f(t)= \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right)$$
\begin{aligned}f(0)&=\cos \left(\frac{\pi}{\sqrt{19-3 \sec 0}}\right) \\
&=\cos \left(\frac{\pi}{\sqrt{19-3}}\right)\\ &=\cos \left(\frac{\pi}{4}\right)\\ &=\frac{\sqrt{2}}{2} \end{aligned}
From (a), (b) since $\lim \limits_{t \rightarrow 0} f(t)=f(0)=\frac{\sqrt{2}}{2},$ the function is continuous at $ t=0$.
