Mathematics

Mathematics

Problem

A function gg has a removable discontinuity at x=1x=1 but is otherwise continuous on (−2,5)(-2,5). After redefining g(1)g(1) to fill the hole, which statement best describes the differentiability of gg at x=1x=1? a) g′(1)g'(1) can never exist since the function was originally undefined at x=1x=1 before the redefinition. b) g′(1)g'(1) exists only if the left-hand and right-hand limits of the difference quotient at x=1x=1 are equal. c) g′(1)g'(1) is guaranteed to exist because redefining g(1)g(1) makes gg continuous at x=1x=1. d) g′(1)g'(1) does not exist because the difference quotient gives the indeterminate form 0/00/0 when x→1x \to 1.
Intuitive explanation

A removable discontinuity means the limit exists as you approach x=1x=1, but g(1)g(1) is either undefined or doesn't match that limit. Once you redefine g(1)g(1) to equal lim⁡x→1g(x)\lim_{x \to 1} g(x), the function becomes continuous there—but continuity alone doesn't guarantee differentiability. The key insight is that differentiability requires the derivative (the slope of the tangent line) to exist, which depends on whether the function's rate of change is well-behaved at that point, not just whether there's no jump.

Intuitive explanation

A removable discontinuity means the limit exists as you approach x=1x=1, but g(1)g(1) is either undefined or doesn't match that limit. Once you redefine g(1)g(1) to equal lim⁡x→1g(x)\lim_{x \to 1} g(x), the function becomes continuous there—but continuity alone doesn't guarantee differentiability. The key insight is that differentiability requires the derivative (the slope of the tangent line) to exist, which depends on whether the function's rate of change is well-behaved at that point, not just whether there's no jump.

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