Exploring the Continuity and Uniform Continuity of ( f(x) frac{1}{x} ) on ( (0,1] )

Understanding the behavior of functions like ( f(x) frac{1}{x} ) is crucial in many areas of mathematics, especially in analyzing their properties within specific intervals. In this article, we will delve into the detailed proof of both the continuity and non-uniform continuity of ( f(x) frac{1}{x} ) on the interval ( (0,1] ).

Step 1: Proving Continuity

To establish that ( f(x) frac{1}{x} ) is continuous at a point ( c ) in the interval ( (0,1] ), we need to verify the epsilon-delta definition of continuity. This requires showing that for every ( epsilon > 0 ), there exists a ( delta > 0 ) such that: [ 0

Step 2: Proving Not Uniformly Continuous

To show that ( f(x) frac{1}{x} ) is not uniformly continuous on ( (0,1] ), we use the definition of uniform continuity. A function ( f ) is uniformly continuous on a set if for every ( epsilon > 0 ), there exists a ( delta > 0 ) such that: [ |x - y|

Let's consider two points ( x frac{1}{n} ) and ( y frac{1}{n 1} ) for large ( n ). Then we have:

[ x - y left| frac{1}{n} - frac{1}{n 1} right| left| frac{n - (n 1)}{nn 1} right| frac{1}{nn 1} ] As ( n to infty ), ( x - y to 0 ).

Now, let's compute ( |f(x) - f(y)| ):

[ f(x) - f(y) frac{1}{frac{1}{n}} - frac{1}{frac{1}{n 1}} n - (n 1) 1 ] For any choice of ( delta ), we can find ( n ) large enough such that ( |x - y|

Conclusion

Therefore, we conclude that ( f(x) frac{1}{x} ) is continuous on ( (0,1] ) but not uniformly continuous on this interval.