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

Both $f(x)=\sin x$ and $g(x)=\cos x$ are continuous at every point $x=c$. The proof is discussed in detail below.
1) $f(x)=\sin x$ To prove that $f$ is continuous at every point $x=c$, we can apply the condition proved in Exercise 69: $$\lim_{h\to0}f(c+h)=f(c)$$ - Examine $\lim_{h\to0}f(c+h)$: $$\lim_{h\to0}f(c+h)=\lim_{h\to0}\sin(c+h)$$ Apply the given identity: $$\lim_{h\to0}f(c+h)=\lim_{h\to0}(\sin h\cos c+\cos h\sin c)$$ $$\lim_{h\to0}f(c+h)=\sin0\cos c+\cos0\sin c$$ $$\lim_{h\to0}f(c+h)=0\cos c+1\sin c=\sin c$$ - Examine $f(c)$: $f(c)=\sin c$ That means $\lim_{h\to c}f(c+h)=f(c)$. $f(x)=\sin x$ is continuous at every point $x=c$. 2) $g(x)=\cos x$ To prove that $g$ is continuous at every point $x=c$, we can apply the condition proved in Exercise 69: $$\lim_{h\to0}g(c+h)=g(c)$$ - Examine $\lim_{h\to0}g(c+h)$: $$\lim_{h\to0}g(c+h)=\lim_{h\to0}\cos(c+h)$$ Apply the given identity: $$\lim_{h\to0}g(c+h)=\lim_{h\to0}(\cos h\cos c-\sin h\sin c)$$ $$\lim_{h\to0}g(c+h)=\cos0\cos c-sin0\sin c$$ $$\lim_{h\to0}g(c+h)=1\cos c-0\sin c=\cos c$$ - Examine $g(c)$: $g(c)=\cos c$ That means $\lim_{h\to c}g(c+h)=g(c)$. $g(x)=\cos x$ is continuous at every point $x=c$.