Answer
$$\lim_{y\to1}\sec(y\sec^2y-\tan^2y-1)=1$$
The function is continuous at $y=1$
Work Step by Step
*Recall Theorem 10:
If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$
$$A=\lim_{y\to1}f(y)=\lim_{y\to1}\sec(y\sec^2y-\tan^2y-1)$$ $$A=\lim_{y\to1}\sec\Big(\frac{y}{\cos^2y}-\frac{\sin^2y}{\cos^2y}-1\Big)$$ $$A=\lim_{y\to1}\sec\Big(\frac{y-\sin^2y-\cos^2y}{\cos^2y}\Big)$$ $$A=\lim_{y\to1}\sec\Big(\frac{y-1}{\cos^2y}\Big)$$
- First, consider the function $y=\sec x=1/\cos x$
As $\cos x\ne0$, or $x\ne\pi/2+k\pi$ $(k\in Z)$, $\lim_{x\to c}\sec x= \sec c$
So function $y=\sec x$ is continuous for all $x\ne\pi/2+k\pi$ $(k\in Z)$
That means we can apply Theorem 10 here as long as $\frac{y-1}{\cos^2y}\ne\pi/2+k\pi$. In detail,
$$A=\sec\Big(\frac{\lim_{y\to1}y-1}{\lim_{y\to1}\cos^2y}\Big)$$ $$A=\sec\Big(\frac{1-1}{\cos^21}\Big)$$ $$A=\sec\Big(\frac{0}{\cos^21}\Big)=\sec0=1$$
- Examine $f(1)$: $$\sec\Big(\frac{1-1}{\cos^21}\Big)=\sec\Big(\frac{0}{\cos^21}\Big)=\sec0=1$$
As $\lim_{y\to1}f(y)=f(1)$, the function is continuous at $y=1$
