Check uniform continuity of functions Pick out the uniformly continuous functions:
$$
\begin{array}{lll}
(a) \quad& f(x) = \cos   x \cos \pi x                   & x \in(0, 1)\\
(b)      & f(x) = \sin   x \cos \pi x                   & x \in(0, 1)\\
(c)      & f(x) = \sin^2 x                              & x \in(0, \infty)\\
(d)      & f(x) = \cos   x \cos \left(\frac\pi x\right) & x \in(0, 1)\\
(e)      & f(x) = \sin   x \cos \left(\frac\pi x\right) & x \in(0, 1)\\
\end{array}
$$
 A: We can solve these by knowing these things

Let $f:(a,b)\to\mathbb R$ be a function that is continuous on the bounded, open interval $(a,b)$. Then the two limits
  $$\lim_{x\to a_+}f(x)\qquad\text{and}\qquad \lim_{x\to b_-}f(x)$$
  exist if and only if $f$ is uniformly continuous on $(a,b)$
Theorem 1.12 in The calculus integral by Brian S. Thomson

and

If a function $f$ has a bounded derivative, it is Lipschitz continuous.
  All Lipschitz continuous functions are uniformly continuous.


$$(a)\quad f(x)=\cos x \cos(\pi x),\quad x\in(0,1)$$
The derivative of this function is
$$f'(x)=-\sin x\cos(\pi x)-\pi \sin(\pi x) \cos x$$
This functions derivative is bounded by $[-\pi-1,\pi+1]$, hence it's Lipschitz continuous.

$$(b)\quad f(x)=\sin x\cos (\pi x),\quad x\in(0,1)$$
The derivative of this function is
$$f'(x)=\cos x\cos(\pi x)-\pi \sin x \sin(\pi x)$$
This functions derivative is bounded by $[-\pi-1,\pi+1]$, hence it's Lipschitz continuous.

$$(c)\quad f(x)=\sin^2 x,\quad x\in(0,\infty)$$
The derivative of this function is
$$f'(x)=\sin(2x)$$
This functions derivative is bounded by $[-1,1]$, hence it's Lipschitz continuous.

$$(d)\quad f(x) = \cos x\cos\left(\frac\pi x\right),\quad x\in(0,1)$$
The limit below is undefined, hence the function is not uniformly continuous
$$\lim_{x\to0_+} \cos x\cos\left(\frac\pi x\right)=\color{gray}{\text{undefined}}$$

$$(d)\quad f(x) = \sin x\cos\left(\frac\pi x\right),\quad x\in(0,1)$$
The limits below are defined, hence the function is uniformly continuous
$$\lim_{x\to0_+} \sin x\cos\left(\frac\pi x\right)=0$$
$$\lim_{x\to1_-} \sin x\cos\left(\frac\pi x\right)=-\sin 1$$
