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

$$\lim_{x\to1}\cos^{-1}(\ln\sqrt x)=\frac{\pi}{2}$$ The function is continuous at $x=1$.
*Recall Theorem 10: If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$ $$A=\lim_{x\to1}f(x)=\lim_{x\to1}\cos^{-1}(\ln\sqrt x)$$ Apply Theorem 10 here: $$A=\cos^{-1}(\ln\sqrt{\lim_{x\to1}x})$$ $$A=\cos^{-1}(\ln\sqrt1)=f(1)$$ $$A=\cos^{-1}(\ln1)$$ $$A=\cos^{-1}0$$ $$A=\frac{\pi}{2}$$ As shown above, $\lim_{x\to1}f(x)=f(1)$, so the function is continuous at $x=1$.