Finding reliable solutions for Elementary Analysis: The Theory of Calculus
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Let ( \epsilon > 0 ) be given. Choose ( \delta = \min(1, 2\epsilon) ). Suppose ( |x-2| < \delta ). Then ( |x-2| < 1 ) implies ( x > 1 ), hence ( \frac1 < 1 ). Now compute: [ \left|\frac1x - \frac12\right| = \frac < \frac2 < \frac2\epsilon2 = \epsilon. ] Thus, ( f(x) = 1/x ) is continuous at ( x=2 ). ( \blacksquare ) 2\epsilon) ). Suppose ( |x-2| <