Check uniform continuity of functions

Question

Pick out the uniformly continuous functions:

$$ \begin{array}{lll} (a) \quad& f(x) = \cos x \cos \pi x & x \in(0, 1)\\ (b) & f(x) = \sin x \cos \pi x & x \in(0, 1)\\ (c) & f(x) = \sin^2 x & x \in(0, \infty)\\ (d) & f(x) = \cos x \cos \left(\frac\pi x\right) & x \in(0, 1)\\ (e) & f(x) = \sin x \cos \left(\frac\pi x\right) & x \in(0, 1)\\ \end{array} $$

Answer 1

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.

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$$(a)\quad f(x)=\cos x \cos(\pi x),\quad x\in(0,1)$$ 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.

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$$(b)\quad f(x)=\sin x\cos (\pi x),\quad x\in(0,1)$$ 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.

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

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$$(d)\quad f(x) = \cos x\cos\left(\frac\pi x\right),\quad x\in(0,1)$$

The limit below is undefined, hence the function is not uniformly continuous

$$\lim_{x\to0_+} \cos x\cos\left(\frac\pi x\right)=\color{gray}{\text{undefined}}$$

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$$(d)\quad f(x) = \sin x\cos\left(\frac\pi x\right),\quad x\in(0,1)$$

The limits below are defined, hence the function is uniformly continuous

$$\lim_{x\to0_+} \sin x\cos\left(\frac\pi x\right)=0$$ $$\lim_{x\to1_-} \sin x\cos\left(\frac\pi x\right)=-\sin 1$$