Worksheet: Formal Definition of a Limit

In this worksheet, we will practice interpreting the formal epsilon–delta definition of a limit and finding the value of delta for a given epsilon.

Q1:

Find the largest 𝛿 > 0 such that if | π‘₯ βˆ’ 5 | < 𝛿 , then | | | 1 π‘₯ βˆ’ 1 5 | | | < 1 1 0 . Give your answer as a fraction.

  • A 2 5 3
  • B 1 2 5 5
  • C 4 7 1 0
  • D 5 3
  • E 5 3 3

Q2:

Find the largest 𝛿 > 0 such that if | π‘₯ βˆ’ 5 | < 𝛿 , then | | | 1 π‘₯ βˆ’ 1 5 | | | < πœ€ . Give your answer as a fraction involving πœ€ .

  • A 2 5 πœ€ + 1 0 5 πœ€ + 1
  • B 5 5 πœ€ + 1
  • C βˆ’ 2 5 πœ€ 5 πœ€ + 1
  • D 2 5 πœ€ 5 πœ€ + 1
  • E βˆ’ 2 5 πœ€ + 1 0 5 πœ€ + 1

Q3:

Find the largest π‘Ž > 0 such that if | π‘₯ βˆ’ π‘Ž | < π‘Ž , then | | | 1 π‘₯ βˆ’ 1 π‘Ž | | | < πœ€ . Give your answer as a fraction involving πœ€ and π‘Ž .

  • A π‘Ž πœ€ + 2 π‘Ž π‘Ž πœ€ + 1 
  • B π‘Ž πœ€ + 1 π‘Ž πœ€ 
  • C π‘Ž πœ€ βˆ’ 2 π‘Ž π‘Ž πœ€ + 1 
  • D π‘Ž πœ€ π‘Ž πœ€ + 1 
  • E π‘Ž πœ€ βˆ’ 1 π‘Ž πœ€ 

Q4:

In the figure, we have the graph 𝑦 = 𝑓 ( π‘₯ ) that is increasing and concave up. Next to ( π‘₯ , 𝑦 ) are the points ( π‘Ž , 𝑦 + π‘˜ ) and ( 𝑏 , 𝑦 βˆ’ π‘˜ ) for a small π‘˜ > 0 .

Which point is closer to π‘₯ along the horizontal axis, π‘Ž or 𝑏 ?

  • A π‘Ž
  • B 𝑏

What is the distance 𝛿 between the nearest point and π‘₯ from your answer above? Give an expression that involves 𝑓 and absolute values.

  • A | | π‘₯ + 𝑓 ( 𝑓 ( π‘₯ ) βˆ’ π‘˜ ) | | βˆ’ 1
  • B | | π‘₯ + 𝑓 ( 𝑓 ( π‘₯ ) + π‘˜ ) | | βˆ’ 1
  • C | | π‘₯ βˆ’ 𝑓 ( 𝑓 ( π‘₯ ) βˆ’ π‘˜ ) | | βˆ’ 1
  • D | | π‘₯ βˆ’ 𝑓 ( 𝑓 ( π‘₯ ) + π‘˜ ) | | βˆ’ 1

Use your answers above to find 𝛿 such that | 𝑒 βˆ’ 1 | < 0 . 1 π‘₯ whenever | π‘₯ βˆ’ 0 | < 𝛿 . Give your answer to 4 decimal places.

Q5:

The figure shows the graph of 𝑓 ( π‘₯ ) = 2 π‘₯ around a point ( π‘₯ , 𝑦 ) where π‘₯ > 0 and 𝑦 = 𝑓 ( π‘₯ ) . Nearby are points ο€Ό π‘Ž , 𝑦 + 1 2  and ο€Ό 𝑏 , 𝑦 βˆ’ 1 2  .

From the graph, looking at the π‘₯ -axis, which of the points π‘Ž and 𝑏 is nearest to π‘₯ ? Let this number be 𝑝 .

  • A π‘Ž
  • B 𝑏

What is π‘Ž in terms of π‘₯ ?

  • A 4 π‘₯ 4 + π‘₯
  • B 2 π‘₯ 4 + π‘₯
  • C 4 + π‘₯ 2 π‘₯
  • D 4 + π‘₯ 4 π‘₯
  • E 4 π‘₯

The claim is that if π‘₯ > 0 and 2 π‘₯ > 1 2 , then | | | 2 π‘₯ βˆ’ 2 π‘₯ | | | < 1 2 βˆ— provided that | π‘₯ βˆ’ π‘₯ | < 𝛿 βˆ— is true for all small 𝛿 . What is the largest such 𝛿 ?

  • A π‘₯ π‘₯ + 2 2
  • B π‘₯ + 4 π‘₯ 2
  • C π‘₯ π‘₯ + 4 2
  • D π‘₯ π‘₯ + 4
  • E π‘₯ + 4 π‘₯

The condition 2 π‘₯ > 1 2 was to ensure truth to the figure. Does the same 𝛿 work if π‘₯ β‰₯ 4 ?

  • Ano
  • Byes

It would appear that all your arguments depended upon the fact that the graph of the function 𝑓 was concave up. Let 𝑓 ( π‘₯ ) = 𝑒 βˆ’ π‘₯ . Find 𝛿 > 0 so that | 𝑓 ( π‘₯ ) βˆ’ 𝑓 ( π‘₯ ) | < 1 4 βˆ— whenever | π‘₯ βˆ’ π‘₯ | < 𝛿 βˆ— . Assume that π‘₯ is positive.

  • A 𝛿 = 2 π‘₯ βˆ’ 4 + ( 4 + 𝑒 ) l n l n βˆ’ π‘₯
  • B 𝛿 = 2 π‘₯ βˆ’ 4 + ( 4 + 𝑒 ) l n l n π‘₯
  • C 𝛿 = ( 4 + 𝑒 ) βˆ’ ( 4 ) l n l n βˆ’ π‘₯
  • D 𝛿 = ( 4 + 𝑒 ) βˆ’ ( 4 ) l n l n π‘₯
  • E 𝛿 = ( 4 + 𝑒 ) + ( 4 ) l n l n π‘₯

Q6:

Sometimes, we do not have concavity to help determine an explicit 𝛿 that shows continuity at a point. The figure shows the graph of

together with its tangent line 𝑦 = π‘₯ + 3 at the inflection point ( 0 , 3 ) .

What is 𝑓 ( π‘₯ ) βˆ’ 1 for π‘₯ > 0 ?

  • A 1 βˆ’ √ 4 9 βˆ’ 1 6 π‘₯ 8
  • B βˆ’ √ 4 π‘₯ βˆ’ 1 1 + 1 2
  • C 1 + √ 4 9 βˆ’ 1 6 π‘₯ 8
  • D √ 4 π‘₯ βˆ’ 1 1 βˆ’ 1 2

What is 𝑓 ( π‘₯ ) βˆ’ 1 for π‘₯ ≀ 0 ?

  • A βˆ’ √ 4 π‘₯ βˆ’ 1 1 + 1 2
  • B √ 4 π‘₯ βˆ’ 1 1 βˆ’ 1 2
  • C 1 + √ 4 9 βˆ’ 1 6 π‘₯ 8
  • D 1 βˆ’ √ 4 9 βˆ’ 1 6 π‘₯ 8

By considering the graph near ( 0 , 3 ) , find the largest 𝛿 so that | 𝑓 ( π‘₯ ) βˆ’ 3 | < πœ€ whenever | π‘₯ | < 𝛿 .

  • A √ 1 6 πœ€ + 1 βˆ’ 1 8
  • B 1 βˆ’ √ 1 6 πœ€ + 1 8
  • C √ 4 πœ€ + 1 βˆ’ 1 2
  • D 1 βˆ’ √ 4 πœ€ + 1 2

Q7:

The graph of 𝑓 ( π‘₯ ) = 5 βˆ’ ( π‘₯ βˆ’ 2 ) 2 is concave down and decreasing when π‘₯ > 2 . We want to find the maximal 𝛿 so that, for a given πœ€ > 0 , it follows that if | π‘₯ βˆ’ 3 | < 𝛿 then | 𝑓 ( π‘₯ ) βˆ’ 𝑓 ( 3 ) | < πœ€ . This 𝛿 will, of course, be a function of πœ€ .

What is the inverse function near π‘₯ = 3 ? Give an expression for 𝑓 ( π‘₯ ) βˆ’ 1 .

  • A 2 + √ 5 + π‘₯
  • B 2 βˆ’ √ 5 βˆ’ π‘₯
  • C 2 βˆ’ √ 5 + π‘₯
  • D 2 + √ 5 βˆ’ π‘₯
  • E 2 + √ π‘₯ βˆ’ 5

Using the concavity of the graph of 𝑓 , determine which point is closer to 3, 𝑓 ( 4 + πœ€ ) βˆ’ 1 or 𝑓 ( 4 βˆ’ πœ€ ) βˆ’ 1 , along the horizontal axis. (Do not evaluate them and consider only small πœ€ .)

  • A 𝑓 ( 4 βˆ’ πœ€ ) βˆ’ 1
  • B 𝑓 ( 4 + πœ€ ) βˆ’ 1

From your answers above, find an expression β€” in terms of πœ€ β€” for the largest 𝛿 so that if | π‘₯ βˆ’ 3 | < 𝛿 then 4 βˆ’ πœ€ < 𝑓 ( π‘₯ ) < 4 + πœ€ .

  • A √ 1 + πœ€ βˆ’ 1
  • B 1 βˆ’ √ 1 βˆ’ πœ€
  • C 2 + √ 1 + πœ€
  • D 2 + √ 1 βˆ’ πœ€

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