Answer
$$\lim_{t\to0}\cos\Big(\frac{\pi}{\sqrt{19-3\sec2t}}\Big)=\frac{\sqrt2}{2}$$
The function is continuous at $t=0$.
Work Step by Step
*Recall Theorem 10:
If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$
$$A=\lim_{t\to0}f(x)=\lim_{t\to0}\cos\Big(\frac{\pi}{\sqrt{19-3\sec2t}}\Big)$$
- With function $y=\cos x$:
For all $x\in R$, $\lim_{x\to c}\cos x=\cos c$, so $y=\cos x$ is continuous on $(-\infty,\infty)$
That means we can apply Theorem 10 here with no conditions. In detail,
$$A=\cos\Big(\frac{\pi}{\sqrt{19-3\lim_{t\to0}\sec2t}}\Big)$$ $$A=\cos\Big(\frac{\pi}{\sqrt{19-3\sec0}}\Big)=f(0)$$ $$A=\cos\Big(\frac{\pi}{\sqrt{19-3\times1}}\Big)=\cos\Big(\frac{\pi}{\sqrt{16}}\Big)=\cos\frac{\pi}{4}$$ $$A=\frac{\sqrt2}{2}$$
As shown above, $\lim_{t\to0}f(t)=f(0)$, so the function is continuous at $t=0$
