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Is $f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$ uniform convergence on $ [0,1]$?

Of course we have $f_n \rightarrow \cos(x)$ but how prove that $$\operatorname{sup} \left| \cos \left( \sqrt{x^2 + \frac{1}{n} }\right) - \cos(x) \right| \rightarrow 0 $$ or $$\operatorname{sup} \left| \cos \left( \sqrt{x^2 + \frac{1}{n} }\right) - \cos(x) \right|\not\rightarrow 0 $$ ?

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Hints:

$$ \cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} $$

$$ \sqrt{x^2+\frac{1}{n}}-x=\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}} $$

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