If the limit of f(x) as x approaches c is 0, then there must exist a number k such that f(k) < 0.001. Is this true or false?
It's a true statement x approaches c, and at the same time f(x) approaches to zero, so if you closen x to c, the f(x) will get closer to 0 think carefully it means f approaches to zero like this: 1...0.1...0.001...0.0001....... 0 so there will be a special number (that you called it k) close to c that if you replace x with it, the f(x) won't be the exact zero but it has closened to zero and will be smaller than 0.001 and if you assume k, closer to c, and then closer, f will be smaller and smaller that it will reach zero in limit
From the definition of the limit, \[\large\lim_{x\to c}f(x)=0\] means there is some \(\delta\) such that \(|x-c|<\delta\) implies that \(|f(x)-0|=|f(x)|<\epsilon\) for all \(\epsilon>0\). Replace \(\delta\) with \(k\) to match the problem statement. \(\epsilon\) can be any number, such as \(0.001\).
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