Question #311001

Let f(x)= cos 1/x is not uniformly continuous on (0, infinity)

Expert's answer

True. For xn=12πn,yn=1π2+2πnx_n=\frac{1}{2\pi n},y_n=\frac{1}{\frac{\pi}{2}+2\pi n}

f(xn)−f(yn)=cos⁡(2πn)−cos⁡(π2+2πn)=1−0=1f\left( x_n \right) -f\left( y_n \right) =\cos \left( 2\pi n \right) -\cos \left( \frac{\pi}{2}+2\pi n \right) =1-0=1

Meanwhile ∣xn−yn∣=∣12πn−1π2+2πn∣=12πn(n+4)→0,n→∞\left| x_n-y_n \right|=\left| \frac{1}{2\pi n}-\frac{1}{\frac{\pi}{2}+2\pi n} \right|=\frac{1}{2\pi n\left( n+4 \right)}\rightarrow 0,n\rightarrow \infty

This contradicts with the definition of uniform continuity.


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