Let f(x)= cos 1/x is not uniformly continuous on (0, infinity)
True. For xn=12πn,yn=1π2+2πnx_n=\frac{1}{2\pi n},y_n=\frac{1}{\frac{\pi}{2}+2\pi n}xn=2πn1,yn=2π+2πn1
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=1f(xn)−f(yn)=cos(2πn)−cos(2π+2πn)=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∣xn−yn∣=∣∣2πn1−2π+2πn1∣∣=2πn(n+4)1→0,n→∞
This contradicts with the definition of uniform continuity.
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