Exercise 2: Which Part of Continuity Fails at x=3? — Possible Solution
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THIS CHAPTER'S OWN THREE-PART DEFINITION
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f is continuous at a when: (1) f(a) is defined, (2) the limit
lim(x->a) f(x) exists, and (3) the two are equal.

CHECKING EACH PART FOR f(x)=(x^2-9)/(x-3) AT x=3
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Part 1 - is f(3) defined? Substituting x=3 directly gives
(9-9)/(3-3) = 0/0, a genuine division by zero. f(3) is NOT
defined. Part 1 FAILS.

Part 2 - does the limit exist? Per Exercise 1, lim(x->3) f(x) = 6,
confirmed both algebraically and numerically from both sides. The
limit DOES exist. Part 2 holds.

Part 3 - are they equal? This part cannot even be evaluated
meaningfully, since f(3) doesn't exist in the first place - there
is nothing for the limit value of 6 to be compared against.

RESULT
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The specific failure is Part 1: f(3) itself is undefined (a
genuine division-by-zero), even though Part 2 (the limit existing)
holds perfectly fine. This is exactly the same "removable
discontinuity" pattern this chapter's own (x^2-4)/(x-2) example
showed - a real hole in the function at exactly one point, even
though the function approaches a perfectly well-defined value there.

WHY THIS WORKS AS AN ANSWER
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Each of the three named conditions is checked individually and
explicitly, rather than just concluding "it's discontinuous"
without identifying which specific requirement is the one that
actually fails, and the result is tied back to this chapter's own
already-established terminology (removable discontinuity) rather
than treated as an unrelated new case.