Which sentence states that the limit of f(x) as x tends to a is L?

• For every epsilon above 0 there is a delta above 0 with 0 < |x - a| < delta giving |f(x) - L| < epsilon
• For every delta there is an epsilon with 0 < |x - a| < delta giving |f(x) - L| < epsilon
• There is one epsilon so that |x - a| < 1 gives |f(x) - L| < epsilon

Let f(x) = 2x + 5. We know the limit of f(x) as x approaches 1 is 7. For the tolerance epsilon = 0.1, what positive value of delta works in the definition? Give your answer as a decimal.

Answer: ______________

Which statement is the correct epsilon-delta meaning of 'the limit of f(x) as x approaches a is L'?

• For every delta there is an epsilon with 0 less than |x - a| less than delta implying |f(x) - L| less than epsilon
• There is one epsilon so that 0 less than |x - a| less than 1 implies |f(x) - L| less than epsilon
• For every epsilon greater than 0 there is a delta greater than 0 so that 0 less than |x - a| less than delta implies |f(x) - L| less than epsilon
• For every epsilon there is a delta so that |x - a| less than delta implies f(x) equals L

Someone claims a limit is L. In the definition, which value is picked before anything else?

• The distance delta around a
• The tolerance epsilon around L
• The value f(a) at the point

Why does the definition of a limit require the statement to work for every epsilon, instead of just one small epsilon?

• Because epsilon must always be smaller than delta
• Because the challenger could demand any level of closeness to L, so a single epsilon leaves most demands unanswered
• Because delta is usually larger than epsilon
• Because the value f(a) is never known

To prove the limit of 3x + 1 as x approaches 1 is 4, it is enough to show that one small epsilon, say 0.001, has a working delta.

Circle one:   True   False

Let f(x) = 4x - 3, and take the limit as x approaches 2, which is 5. A classmate is challenged with epsilon = 0.02. Which delta gives a complete answer?

• 0.08
• 0.005
• 0.02
• 0.5

Showing one small epsilon with a working delta proves a limit.

Circle one:   True   False

To prove 3x + 1 tends to 7 as x tends to 2 for an arbitrary epsilon, which delta completes the proof?

• 3 times epsilon
• epsilon / 3
• epsilon + 3

Let f(x) = x squared, with limit 0 as x tends to 0. For epsilon = 0.25, give the largest delta that works.

Answer: ______________