Is $f(x)=\frac{\cos x}x$ uniformly continuous in $(0,1)$?

Question

Show that $$f(x)=\frac{\cos x}{x}$$ is not uniformly continuous on $(0,1)$

My attempt:

Here, $\lim_{x \mapsto 0} \frac{\cos x}{x}$ does't exist, we cannot continuously extend $f$ on $[0,1]$. So $f$ is not uniformly continuous on $(0,1)$.

My interest is to show this by sequential criterion. Is my following attempt correct ?

Take $$ x_n=\frac{1}{n} \quad \text{and} \quad y_n=\frac{1}{n+1}. $$ Then $\vert x_n - y_n \vert \rightarrow 0$. But $$ \vert\, f(x_n) - f(y_n) \vert = \left\vert \frac{\cos 1/n}{1/n} -\frac{\cos 1/(n+1)}{1/(n+1)}\right\vert \rightarrow1 (?) $$

Any help?

Answer 1

Correct. It does tend to 1 $$ \frac{\cos\big(\frac{1}{n+1}\big)}{\frac{1}{n+1}}-\frac{\cos\big(\frac{1}{n+1}\big)}{\frac{1}{n}}=(n+1)\cos\Big(\frac{1}{n+1}\Big)-n\cos\Big(\frac{1}{n}\Big) \\ =\cos\Big(\frac{1}{n+1}\Big)+n\left(\cos\Big(\frac{1}{n+1}\Big)-\cos\Big(\frac{1}{n}\Big)\right) =\cos\Big(\frac{1}{n+1}\Big)-n \left(\frac{1}{n+1}-\frac{1}{n}\right)\sin \xi_n \\=\cos\Big(\frac{1}{n+1}\Big)+\frac{\sin\xi_n}{n+1}\to 1, $$ as $n\to \infty$.

Answer 2

The first approach (the function cannot be continuously extended to $[0,1]$) is correct. The other one no, of course, since the equality $$\left\vert \frac{\cos 1/n}{1/n} -\frac{\cos 1/(n+1)}{1/(n+1)}\right\vert=1$$is false.

However, it is indeed true that$$\lim_{n\to\infty}\left\vert \frac{\cos 1/n}{1/n} -\frac{\cos 1/(n+1)}{1/(n+1)}\right\vert=1,$$and that's enough for your proof.