A trigonometric polynomial.

$f(x)= \sin x + \frac 12 \cos 2x$ .

This function is defined on $\mathbb{R}$ and is periodic with period equal to $2\pi$ ; therefore we can study it on $[0,2 \pi ]$ .

Because of the periodicity, the function $f$ has no limit at $\pm \infty$ . Moreover, there is no other open endpoint of the domain and the function is continuous on $\mathbb{R}$ , therefore we have no limit to compute.

The function is given by a trigonometric polynomial, thus it is differentiable on $\mathbb{R}$ , whence on $[0,2 \pi ]$ .

$$\forall x \in [0,2 \pi ], \; f'(x)=\cos x - \sin x.$$

Where does the first derivative vanish?

$$f'(x)=0 \Longleftrightarrow \cos x = \sin x \Longleftrightarrow x=\frac {\pi}{4} + 2k \pi, \; k \in \mathbb{Z}.$$

Figure: The graph of $f(x)= \sin x + \frac 12 \cos 2x$ .