Nadia says x squared is continuous at 1 because whenever a sequence converging to 1 is squared, the outputs converge to 1. Is her use of the sequential criterion correct?

Circle one:   True   False

Which function is continuous but not uniformly continuous on its domain?

• f(x) equal to x on [0, 1]
• f(x) equal to 1 over x on (0, 1)
• The constant function f(x) equal to 5

For f(x) equal to 3x at 2, which delta answers a given epsilon?

• Delta equal to epsilon over 3
• Delta equal to epsilon
• Delta equal to 3

Nadia says x squared is continuous at 1 because whenever a sequence xn converges to 1, xn squared converges to 1. Is her use of the sequential criterion correct?

Circle one:   True   False

Which function is continuous but not uniformly continuous (on its given domain)?

• f(x) = 1/x on (0, 1)
• f(x) = x on [0, 1]
• The constant function f(x) = 5
• f(x) = x squared on [0, 1]

The function f(x) equal to x on (0, 1) attains both a maximum and a minimum.

Circle one:   True   False

Which example shows the closedness hypothesis in the extreme value theorem cannot be dropped?

• f(x) = x on the open interval (0, 1), which attains neither maximum nor minimum
• f(x) = x on [0, 1]
• f(x) = x squared on [0, 1]
• A constant function on (0, 1)

For f(x) = 3x at c = 2, which delta works for a given epsilon in the epsilon-delta definition?

• delta = epsilon / 3
• delta = epsilon
• delta = 3
• delta = epsilon squared

A student argues 1 over x on (0, 1) must be uniformly continuous because it is continuous at each point. Where is the flaw?

• Continuity at each point allows a fresh delta per point, while uniformity needs one shared delta
• 1 over x is actually discontinuous on (0, 1)
• Uniformity only concerns unbounded functions

Why is f(x) = 1/x not uniformly continuous on (0, 1)?

• Points near 0 need ever smaller deltas, so no one delta works everywhere
• It is discontinuous at some point of (0, 1)
• An unbounded function cannot be continuous
• Uniform continuity only applies to polynomials