Chapter 1: Limits
Section 1.6: Continuity
Discuss the continuity of fx=1/x−1 at x=1.
The point x=1 is the left endpoint of the domain of f, and at this point, f is discontinuous; the discontinuity is not removable. Here's how these conclusions are reached.
The graph of f in Figure 1.6.4(a) suggests its domain is the open interval 1,∞.
Since f1 is undefined (division by zero), x=1 is not in the domain of f, and likewise any x<1, but these because the square root is real only where x−1>0.
Because f1 is undefined, f cannot be continuous at x=1.
Since x=1 is the equation of a vertical asymptote for f (because limx→1+1/x−1 = ∞), f1 cannot be defined to make its extension to 1,∞ continuous.
Figure 1.6.4(a) Graph of fx=1/x−1
A function cannot be continuous at a point where there is a vertical asymptote.
Such a discontinuity is not removable.
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