Question

find the limits. is the function continuous at the point being approached?
\\(\lim_{x\to\frac{7\pi}{16}}\cos(8x - \cos(8x))\\)
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function is continuous at (x = \frac{7\pi}{16}) because it is a composition of two continuous functions and (\lim_{x\to\frac{7\pi}{16}}\cos(8x - \cos(8x))=) (simplify your answer.)
b. the function is not continuous at (x = \frac{7\pi}{16}) because (\cos x) is not a continuous function.
c. the function is not continuous at (x = \frac{7\pi}{16}) because (8x) is not a continuous function.

Explanation:

Step1: Recall continuity of cosine function

The cosine function $y = \cos(u)$ is continuous for all real - valued $u$. Also, the linear function $y = 8x$ and the cosine function $y=\cos(8x)$ are continuous for all real $x$. The composition of continuous functions is continuous.

Step2: Substitute $x=\frac{7\pi}{16}$ into the function

Let $u = 8x-\cos(8x)$. First, find the value of $u$ when $x = \frac{7\pi}{16}$.
\[

$$\begin{align*} 8x&=8\times\frac{7\pi}{16}=\frac{7\pi}{2}\\ \cos(8x)&=\cos(\frac{7\pi}{2}) = 0 \end{align*}$$

\]
Then $u=8x - \cos(8x)=\frac{7\pi}{2}-0=\frac{7\pi}{2}$.
Now, $\lim_{x
ightarrow\frac{7\pi}{16}}\cos(8x - \cos(8x))=\cos(\frac{7\pi}{2})=0$.

Answer:

A. The function is continuous at $x = \frac{7\pi}{16}$ because it is a composition of two continuous functions and $\lim_{x
ightarrow\frac{7\pi}{16}}\cos(8x - \cos(8x)) = 0$