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Worksheet: Continuity of a Function

Q1:

Discuss the continuity of the function 𝑓 at π‘₯ = βˆ’ 2 , given

  • A The function is discontinuous at π‘₯ = βˆ’ 2 because 𝑓 ( βˆ’ 2 ) is undefined.
  • B The function is discontinuous at π‘₯ = βˆ’ 2 because l i m π‘₯ β†’ βˆ’ 2 𝑓 ( π‘₯ ) does not exist.
  • C The function is discontinuous at π‘₯ = βˆ’ 2 because 𝑓 ( βˆ’ 2 ) β‰  𝑓 ( π‘₯ ) l i m π‘₯ β†’ βˆ’ 2 .
  • D The function is continuous at π‘₯ = βˆ’ 2 .

Q2:

Discuss the continuity of the function 𝑓 at π‘₯ = 2 , given

  • A The function is discontinuous at π‘₯ = 2 because 𝑓 ( 2 ) is undefined.
  • B The function is discontinuous at π‘₯ = 2 because l i m π‘₯ β†’ 2 𝑓 ( π‘₯ ) does not exist.
  • C The function is discontinuous at π‘₯ = 2 because 𝑓 ( 2 ) β‰  𝑓 ( π‘₯ ) l i m π‘₯ β†’ 2 .
  • D The function is continuous at π‘₯ = 2 .

Q3:

Find the values of π‘Ž and 𝑏 that make the function 𝑓 continuous at π‘₯ = βˆ’ 1 and π‘₯ = βˆ’ 6 , given that

  • A π‘Ž = βˆ’ 1 2 7 , 𝑏 = βˆ’ 6 1 7
  • B π‘Ž = 3 7 5 , 𝑏 = 1 2 5
  • C π‘Ž = βˆ’ 6 1 7 , 𝑏 = βˆ’ 1 2 7
  • D π‘Ž = 1 2 5 , 𝑏 = 3 7 5
  • E π‘Ž = 1 2 5 , 𝑏 = βˆ’ 2 5 7 5

Q4:

Consider the function

What is 𝑓 ( 0 ) ?

What is l i m π‘₯ β†’ 0 βˆ’ 𝑓 ( π‘₯ ) ?

What is l i m π‘₯ β†’ 0 + 𝑓 ( π‘₯ ) ?

What type of discontinuity does the function 𝑓 have at π‘₯ = 0 ?

  • AThe function 𝑓 has a removable discontinuity at π‘₯ = 0 .
  • BThe function 𝑓 has an essential discontinuity at π‘₯ = 0 .
  • CThe function 𝑓 does not have a discontinuity at π‘₯ = 0 .
  • DThe function 𝑓 has a jump discontinuity at π‘₯ = 0 .

Q5:

Determine the value of π‘Ž that makes 𝑓 continuous at π‘₯ = 0 , given

  • A βˆ’ 1 0
  • B 5 3
  • C 3 5 3
  • D βˆ’ 6 5 3
  • E βˆ’ 5 3

Q6:

Discuss the continuity of the function 𝑓 at π‘₯ = 0 , given

  • AThe function is discontinuous at π‘₯ = 0 because l i m π‘₯ β†’ 0 𝑓 ( π‘₯ ) does not exist.
  • BThe function is discontinuous at π‘₯ = 0 because 𝑓 ( 0 ) is undefined.
  • CThe function is continuous at π‘₯ = 0 .
  • DThe function is discontinuous at π‘₯ = 0 because l i m π‘₯ β†’ 0 𝑓 ( π‘₯ ) β‰  𝑓 ( 0 ) .

Q7:

The function is continuous at π‘₯ = 0 . Determine the possible values of π‘Ž .

  • A √ 3 , βˆ’ √ 3
  • B √ 3
  • C 2 , βˆ’ 2
  • D 3 , βˆ’ 3

Q8:

Determine the value of π‘Ž that makes the function 𝑓 continuous at π‘₯ = πœ‹ 4 , given

  • A βˆ’ 1 6
  • B 1 2
  • C2
  • D 1 6

Q9:

Find the value of π‘˜ that makes the function 𝑓 continuous at π‘₯ = πœ‹ 4 , given that

  • A βˆ’ 6 5
  • B βˆ’ 2
  • C βˆ’ 6
  • D βˆ’ 2 5

Q10:

Discuss the continuity of the function 𝑓 at π‘₯ = πœ‹ 2 , given

  • AThe function is discontinuous at π‘₯ = πœ‹ 2 because l i m π‘₯ β†’ πœ‹ 2 𝑓 ( π‘₯ ) does not exist.
  • BThe function is discontinuous at π‘₯ = πœ‹ 2 because 𝑓 ο€» πœ‹ 2  is undefined.
  • CThe function is discontinuous at π‘₯ = πœ‹ 2 because l i m π‘₯ β†’ πœ‹ 2 𝑓 ( π‘₯ ) β‰  𝑓 ο€» πœ‹ 2  .
  • DThe function is continuous at π‘₯ = πœ‹ 2 .
  • EThe function is discontinuous on ℝ .

Q11:

Given 𝑓 ( π‘₯ ) = π‘₯ + π‘₯ βˆ’ 2 π‘₯ βˆ’ 1  , if possible or necessary, define 𝑓 ( 1 ) so that 𝑓 is continuous at π‘₯ = 1 .

  • ANo value of 𝑓 ( 1 ) will make 𝑓 continuous because l i m  β†’  𝑓 ( π‘₯ ) does not exist.
  • BThe function is already continuous at π‘₯ = 1 .
  • CThe function cannot be made continuous at π‘₯ = 1 because 𝑓 ( 1 ) is undefined.
  • D 𝑓 ( 1 ) = 3 makes 𝑓 continuous at π‘₯ = 1 .

Q12:

Given 𝑓 ( π‘₯ ) = π‘₯ βˆ’ π‘₯ βˆ’ 6 π‘₯ βˆ’ 3  , if possible or necessary, define 𝑓 ( 3 ) so that 𝑓 is continuous at π‘₯ = 3 .

  • ANo value of 𝑓 ( 3 ) will make 𝑓 continuous because l i m  β†’  𝑓 ( π‘₯ ) does not exist.
  • BThe function is already continuous at π‘₯ = 3 .
  • CThe function cannot be made continuous at π‘₯ = 3 because 𝑓 ( 3 ) is undefined.
  • D 𝑓 ( 3 ) = 5 makes 𝑓 continuous at π‘₯ = 3 .

Q13:

Given 𝑓 ( π‘₯ ) = π‘₯ βˆ’ 3 π‘₯ βˆ’ 2 8 π‘₯ βˆ’ 7  , if possible or necessary, define 𝑓 ( 7 ) so that 𝑓 is continuous at π‘₯ = 7 .

  • ANo value of 𝑓 ( 7 ) will make 𝑓 continuous because l i m  β†’  𝑓 ( π‘₯ ) does not exist.
  • BThe function is already continuous at π‘₯ = 7 .
  • CThe function cannot be made continuous at π‘₯ = 7 because 𝑓 ( 7 ) is undefined.
  • D 𝑓 ( 7 ) = 1 1 makes 𝑓 continuous at π‘₯ = 7 .

Q14:

Given If possible or necessary, define 𝑓 ( βˆ’ 8 ) so that 𝑓 is continuous at π‘₯ = βˆ’ 8 .

  • A 𝑓 ( βˆ’ 8 ) = 6 would make 𝑓 continuous at π‘₯ = βˆ’ 8 .
  • BThe function is already continuous at π‘₯ = βˆ’ 8 .
  • C 𝑓 ( βˆ’ 8 ) = 0 would make 𝑓 continuous at π‘₯ = βˆ’ 8 .
  • DThe function cannot be made continuous at π‘₯ = βˆ’ 8 because l i m π‘₯ β†’ βˆ’ 8 𝑓 ( π‘₯ ) does not exist.

Q15:

Setting 𝑓 ( π‘Ž ) = 5 4 and 𝑓 ( π‘₯ ) = π‘₯ βˆ’ π‘Ž π‘₯ βˆ’ π‘Ž 6 6 3 3 when π‘₯ β‰  π‘Ž makes 𝑓 continuous at π‘₯ = π‘Ž . Determine π‘Ž .

  • A2
  • B 1 3
  • C 1 2
  • D3

Q16:

Setting 𝑓 ( π‘Ž ) = 1 4 and 𝑓 ( π‘₯ ) = π‘₯ βˆ’ π‘Ž π‘₯ βˆ’ π‘Ž 7 7 4 4 when π‘₯ β‰  π‘Ž makes 𝑓 continuous at π‘₯ = π‘Ž . Determine π‘Ž .

  • A 7 4
  • B 1 2
  • C 4 7
  • D2

Q17:

Given 𝑓 ( π‘₯ ) = π‘₯ βˆ’ 6 4 π‘₯ + π‘₯ βˆ’ 2 0   , if possible or necessary, define 𝑓 ( 4 ) so that 𝑓 is continuous at π‘₯ = 4 .

  • ANo value of 𝑓 ( 4 ) will make 𝑓 continuous because l i m  β†’ οŠͺ 𝑓 ( π‘₯ ) does not exist.
  • BThe function is already continuous at π‘₯ = 4 .
  • CThe function cannot be made continuous at π‘₯ = 4 because 𝑓 ( 4 ) is undefined.
  • D 𝑓 ( 4 ) = 1 6 3 makes 𝑓 continuous at π‘₯ = 4 .

Q18:

Given 𝑓 ( π‘₯ ) = π‘₯ βˆ’ 7 2 9 π‘₯ βˆ’ 8 1   , if possible or necessary, define 𝑓 ( 9 ) so that 𝑓 is continuous at π‘₯ = 9 .

  • ANo value of 𝑓 ( 9 ) will make 𝑓 continuous because l i m  β†’  𝑓 ( π‘₯ ) does not exist.
  • BThe function is already continuous at π‘₯ = 9 .
  • CThe function cannot be made continuous at π‘₯ = 9 because 𝑓 ( 9 ) is undefined.
  • D 𝑓 ( 9 ) = 2 7 2 makes 𝑓 continuous at π‘₯ = 9 .

Q19:

Discuss the continuity of the function 𝑓 at π‘₯ = 5 given

  • A The function is continuous at π‘₯ = 5 .
  • B The function is discontinuous at π‘₯ = 5 because l i m π‘₯ β†’ 5 𝑓 ( π‘₯ ) β‰  𝑓 ( 5 ) .
  • C The function is discontinuous at π‘₯ = 5 because 𝑓 ( 5 ) is undefined.
  • D The function is discontinuous at π‘₯ = 5 because l i m π‘₯ β†’ 5 𝑓 ( π‘₯ ) does not exist.

Q20:

Find the value of π‘Ž that makes 𝑓 continuous at π‘₯ = 3 , given that

Q21:

Find the value of π‘Ž that makes 𝑓 continuous at π‘₯ = 2 , given that

Q22:

Find the value of π‘Ž that makes 𝑓 continuous at π‘₯ = 7 , given that

Q23:

Consider the function

What is 𝑓 ( 0 ) ?

What is l i m π‘₯ β†’ 0 βˆ’ 𝑓 ( π‘₯ ) ?

  • AThe limit does not exist.
  • B βˆ’ ∞
  • C + ∞
  • D0
  • E1

What is l i m π‘₯ β†’ 0 + 𝑓 ( π‘₯ ) ?

  • A1
  • B0
  • C + ∞
  • DThe limit does not exist.
  • E βˆ’ ∞

What type of discontinuity does the function 𝑓 have at π‘₯ = 0 ?

  • AThe function 𝑓 has an essential discontinuity at π‘₯ = 0 .
  • BThe function 𝑓 has a removable discontinuity at π‘₯ = 0 .
  • CThe function 𝑓 has a jump discontinuity at π‘₯ = 0 .
  • DThe function 𝑓 does not have a discontinuity at π‘₯ = 0 .

Q24:

Suppose What can be said of the continuity of 𝑓 at π‘₯ = βˆ’ 1 ?

  • A The function is discontinuous at π‘₯ = βˆ’ 1 because l i m π‘₯ β†’ βˆ’ 1 𝑓 ( π‘₯ ) does not exist.
  • B The function is continuous on ℝ .
  • C The function is discontinuous at π‘₯ = βˆ’ 1 because 𝑓 ( βˆ’ 1 ) is undefined.
  • D The function is continuous at π‘₯ = βˆ’ 1 .
  • E The function is discontinuous at π‘₯ = βˆ’ 1 because 𝑓 ( βˆ’ 1 ) β‰  𝑓 ( π‘₯ ) l i m π‘₯ β†’ βˆ’ 1 .