Question #311001

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

1
Expert's answer
2022-03-17T06:30:15-0400

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)=10=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 xnyn=12πn1π2+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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