Lesson Worksheet: Continuity at a Point Mathematics • Higher Education

In this worksheet, we will practice checking the continuity of a function at a given point.

Q1:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=โˆ’2, given ๐‘“(๐‘ฅ)=๏ฒ๐‘ฅ+8๐‘ฅโˆ’4๐‘ฅโ‰ โˆ’2,โˆ’3๐‘ฅ=โˆ’2.๏Šฉ๏Šจifif

  • AThe function is discontinuous at ๐‘ฅ=โˆ’2 because lim๏—โ†’๏Šฑ๏Šจ๐‘“(๐‘ฅ) does not exist.
  • BThe function is continuous at ๐‘ฅ=โˆ’2.
  • CThe function is discontinuous at ๐‘ฅ=โˆ’2 because ๐‘“(โˆ’2)โ‰ ๐‘“(๐‘ฅ)lim๏—โ†’๏Šฑ๏Šจ.
  • DThe function is discontinuous at ๐‘ฅ=โˆ’2 because ๐‘“(โˆ’2) is undefined.

Q2:

Given ๐‘“(๐‘ฅ)=๏ญโˆ’7๐‘ฅ+8๐‘ฅ<โˆ’8,๐‘ฅ+2๐‘ฅ+4๐‘ฅ>โˆ’8.ifif๏Šฉ If possible or necessary, define ๐‘“(โˆ’8) so that ๐‘“ is continuous at ๐‘ฅ=โˆ’8.

  • AThe function is already continuous at ๐‘ฅ=โˆ’8.
  • B๐‘“(โˆ’8)=0 would make ๐‘“ continuous at ๐‘ฅ=โˆ’8.
  • C๐‘“(โˆ’8)=6 would make ๐‘“ continuous at ๐‘ฅ=โˆ’8.
  • DThe function cannot be made continuous at ๐‘ฅ=โˆ’8 because lim๏—โ†’๏Šฑ๏Šฎ๏Žฉ๐‘“(๐‘ฅ)โ‰ lim๏—โ†’๏Šฑ๏Šฎ๏Žช๐‘“(๐‘ฅ).

Q3:

Suppose ๐‘“(๐‘ฅ)=โŽงโŽจโŽฉ๐‘ฅ+1๐‘ฅโˆ’1,๐‘ฅ<โˆ’1,62๐‘ฅโˆ’10,๐‘ฅโ‰ฅโˆ’1.๏Šฉ๏Šฌ What can be said of the continuity of ๐‘“ at ๐‘ฅ=โˆ’1?

  • AThe function is continuous on โ„.
  • BThe function is discontinuous at ๐‘ฅ=โˆ’1 because ๐‘“(โˆ’1) is undefined.
  • CThe function is continuous at ๐‘ฅ=โˆ’1.
  • DThe function is discontinuous at ๐‘ฅ=โˆ’1 because lim๏—โ†’๏Šฑ๏Šง๐‘“(๐‘ฅ) does not exist.
  • EThe function is discontinuous at ๐‘ฅ=โˆ’1 because ๐‘“(โˆ’1)โ‰ ๐‘“(๐‘ฅ)lim๏—โ†’๏Šฑ๏Šง.

Q4:

Find the values of ๐‘Ž and ๐‘ that make the function ๐‘“ continuous at ๐‘ฅ=โˆ’1 and ๐‘ฅ=โˆ’6, given that ๐‘“(๐‘ฅ)=๏ณ3๐‘ฅ+11,๐‘ฅโ‰คโˆ’6,๐‘Ž๐‘ฅ+๐‘,โˆ’6<๐‘ฅ<โˆ’1,โˆ’5๐‘ฅ+10,๐‘ฅโ‰ฅโˆ’1.๏Šจ

  • A๐‘Ž=125, ๐‘=375
  • B๐‘Ž=375, ๐‘=125
  • C๐‘Ž=โˆ’617, ๐‘=โˆ’127
  • D๐‘Ž=โˆ’127, ๐‘=โˆ’617
  • E๐‘Ž=125, ๐‘=โˆ’2575

Q5:

Given ๐‘“(๐‘ฅ)=๐‘ฅ+๐‘ฅโˆ’2๐‘ฅโˆ’1๏Šจ, if possible or necessary, define ๐‘“(1) so that ๐‘“ is continuous at ๐‘ฅ=1.

  • AThe function is already continuous at ๐‘ฅ=1.
  • BNo value of ๐‘“(1) will make ๐‘“ continuous because lim๏—โ†’๏Šง๐‘“(๐‘ฅ) does not exist.
  • C๐‘“(1)=3 makes ๐‘“ continuous at ๐‘ฅ=1.
  • DThe function cannot be made continuous at ๐‘ฅ=1 because ๐‘“(1) is undefined.

Q6:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=5 given ๐‘“(๐‘ฅ)=โŽงโŽชโŽจโŽชโŽฉ8๐‘ฅ+1๐‘ฅโ‰ค5,๐‘ฅโˆ’25๐‘ฅโˆ’125๐‘ฅ>5.ifif๏Šจ๏Šฉ

  • AThe function is discontinuous at ๐‘ฅ=5 because lim๏—โ†’๏Šซ๐‘“(๐‘ฅ) does not exist.
  • BThe function is discontinuous at ๐‘ฅ=5 because ๐‘“(5) is undefined.
  • CThe function is discontinuous at ๐‘ฅ=5 because lim๏—โ†’๏Šซ๐‘“(๐‘ฅ)โ‰ ๐‘“(5).
  • DThe function is continuous at ๐‘ฅ=5.

Q7:

Find the value of ๐‘Ž that makes ๐‘“ continuous at ๐‘ฅ=3, given that ๐‘“(๐‘ฅ)=๏ฒ๐‘ฅ+๐‘ฅ(๐‘Žโˆ’3)โˆ’3๐‘Ž๐‘ฅโˆ’3๐‘ฅโ‰ 3,7๐‘ฅ+6๐‘ฅ=3.๏Šจifif

Q8:

Setting ๐‘“(๐‘Ž)=54 and ๐‘“(๐‘ฅ)=๐‘ฅโˆ’๐‘Ž๐‘ฅโˆ’๐‘Ž๏Šฌ๏Šฌ๏Šฉ๏Šฉ when ๐‘ฅโ‰ ๐‘Ž makes ๐‘“ continuous at ๐‘ฅ=๐‘Ž. Determine ๐‘Ž.

  • A12
  • B3
  • C13
  • D2

Q9:

Determine the value of ๐‘Ž that makes ๐‘“ continuous at ๐‘ฅ=0, given ๐‘“(๐‘ฅ)=๏ฑโˆ’56๐‘ฅโˆ’57๐‘ฅ3๐‘ฅ,๐‘ฅโ‰ 0,๐‘Ž,๐‘ฅ=0.sintan

  • Aโˆ’653
  • B353
  • Cโˆ’10
  • Dโˆ’53
  • E53

Q10:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=0, given ๐‘“(๐‘ฅ)=๏ฑ๐‘ฅ5๐‘ฅ,๐‘ฅโ‰ 0,5,๐‘ฅ=0.sin

  • AThe function is discontinuous at ๐‘ฅ=0 because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ) does not exist.
  • BThe function is discontinuous at ๐‘ฅ=0 because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ)โ‰ ๐‘“(0).
  • CThe function is continuous at ๐‘ฅ=0.
  • DThe function is discontinuous at ๐‘ฅ=0 because ๐‘“(0) is undefined.

Q11:

The function ๐‘“(๐‘ฅ)=๏ฒ7|๐‘ฅ|๐‘ฅ+17,๐‘ฅ<0,๐‘Ž+9๐‘ฅ,๐‘ฅโ‰ฅ0๏Šจcos is continuous at ๐‘ฅ=0. Determine the possible values of ๐‘Ž.

  • A3,โˆ’3
  • Bโˆš3,โˆ’โˆš3
  • Cโˆš3
  • D2,โˆ’2

Q12:

Determine the value of ๐‘Ž that makes the function ๐‘“ continuous at ๐‘ฅ=๐œ‹4, given ๐‘“(๐‘ฅ)=โŽงโŽจโŽฉ2๐‘ฅ+9๐‘ฅ4+4๐‘ฅ,๐‘ฅโ‰ ๐œ‹4,3๐‘Ž,๐‘ฅ=๐œ‹4.sintansin

  • A12
  • Bโˆ’16
  • C2
  • D16

Q13:

Find the value of ๐‘˜ that makes the function ๐‘“ continuous at ๐‘ฅ=๐œ‹4, given that ๐‘“(๐‘ฅ)=โŽงโŽจโŽฉโˆ’62๐‘ฅ+4๐‘ฅ2๐‘ฅ+2,๐‘ฅโ‰ ๐œ‹4,5๐‘˜,๐‘ฅ=๐œ‹4.sintansin

  • Aโˆ’2
  • Bโˆ’25
  • Cโˆ’65
  • Dโˆ’6

Q14:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=๐œ‹2, given ๐‘“(๐‘ฅ)=๏ดโˆ’7๐‘ฅ+7๐‘ฅ,๐‘ฅโ‰ค๐œ‹2,62๐‘ฅโˆ’1,๐‘ฅ>๐œ‹2.sincoscos

  • AThe function is discontinuous at ๐‘ฅ=๐œ‹2 because ๐‘“๏€ป๐œ‹2๏‡ is undefined.
  • BThe function is continuous at ๐‘ฅ=๐œ‹2.
  • CThe function is discontinuous on โ„.
  • DThe function is discontinuous at ๐‘ฅ=๐œ‹2 because lim๏—โ†’๏‘ฝ๏Žก๐‘“(๐‘ฅ) does not exist.
  • EThe function is discontinuous at ๐‘ฅ=๐œ‹2 because lim๏—โ†’๏‘ฝ๏Žก๐‘“(๐‘ฅ)โ‰ ๐‘“๏€ป๐œ‹2๏‡.

Q15:

Determine the value of ๐‘Ž that makes the function ๐‘“ continuous at ๐‘ฅ=0, given that ๐‘“(๐‘ฅ)=๏ฒ68๐‘ฅ8๐‘ฅ2๐‘ฅ๐‘ฅโ‰ 0,๐‘Žโˆ’6๐‘ฅ=0.sintanifif๏Šจ

  • A30
  • B198
  • C24
  • D9

Q16:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=๐œ‹2, given ๐‘“(๐‘ฅ)=๏ด8+7๐‘ฅ๐‘ฅ<๐œ‹2,7+5๐‘ฅ๐‘ฅโ‰ฅ๐œ‹2.cosifsinif

  • AThe function is discontinuous at ๐‘ฅ=๐œ‹2 because lim๏—โ†’๏‘ฝ๏Žก๐‘“(๐‘ฅ)โ‰ ๐‘“๏€ป๐œ‹2๏‡.
  • BThe function is discontinuous at ๐‘ฅ=๐œ‹2 because ๐‘“๏€ป๐œ‹2๏‡ is undefined.
  • CThe function is continuous at ๐‘ฅ=๐œ‹2.
  • DThe function is discontinuous at ๐‘ฅ=๐œ‹2 because lim๏—โ†’๏‘ฝ๏Žก๐‘“(๐‘ฅ) does not exist.

Q17:

Let ๐‘“(๐‘ฅ)=โŽงโŽจโŽฉ๐œ‹๐‘ฅ5๐‘ฅ,๐‘ฅ<0,๐œ‹๏”๐‘Ž+๏€ผ6๐œ‹5๐‘ฅ๏ˆ๏ ,๐‘ฅโ‰ฅ0.sincos Find all values of ๐‘Ž that make ๐‘“ continuous at ๐‘ฅ=0.

  • A15
  • B2๐œ‹5โˆ’1
  • C65
  • Dโˆ’45

Q18:

Let ๐‘“(๐‘ฅ)=๏ฑโˆ’7๐‘ฅ+78๐‘˜๐‘ฅ,๐‘ฅโ‰ 0,โˆ’24๐‘ฅ,๐‘ฅ=0.cossin Find all values of ๐‘˜ that make ๐‘“ continuous at ๐‘ฅ=0.

  • A๐‘˜โˆˆโ„๏Šฐ
  • B0
  • C๐‘˜โˆˆโ„
  • D๐‘˜โˆˆโ„โˆ’{0}

Q19:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=โˆ’7, given that ๐‘“(๐‘ฅ)=๏ญ|๐‘ฅ+7|,๐‘ฅโ‰คโˆ’2,โˆ’๐‘ฅ+3,๐‘ฅ>โˆ’2.

  • AThe function is discontinuous at ๐‘ฅ=โˆ’7 because ๐‘“(โˆ’7)โ‰ ๐‘“(๐‘ฅ)lim๏—โ†’๏Šฑ๏Šญ.
  • BThe function is discontinuous at ๐‘ฅ=โˆ’7 because ๐‘“(โˆ’7) is undefined.
  • CThe function is continuous at ๐‘ฅ=โˆ’7.
  • DThe function is continuous at all points in โ„โˆ’{โˆ’7}.
  • EThe function is discontinuous at ๐‘ฅ=โˆ’7 because lim๏—โ†’๏Šฑ๏Šญ๐‘“(๐‘ฅ) does not exist.

Q20:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=0, given that ๐‘“(๐‘ฅ)=โŽงโŽชโŽจโŽชโŽฉ6๐‘ฅโˆ’๐‘ฅ๐‘ฅ4๐‘ฅ,๐‘ฅ<0,๐‘ฅ+5๐‘ฅ+4,๐‘ฅโ‰ฅ0.๏Šจ๏Šจsintan

  • AThe function is discontinuous on โ„.
  • BThe function is discontinuous at ๐‘ฅ=0 because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ) does not exist.
  • CThe function is continuous at ๐‘ฅ=0.
  • DThe function is discontinuous at ๐‘ฅ=0 because ๐‘“(0) is undefined.
  • EThe function is discontinuous at ๐‘ฅ=0 because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ)โ‰ ๐‘“(0).

Q21:

Find the value of ๐‘˜ which makes the function ๐‘“ continuous at ๐‘ฅ=0, given ๐‘“(๐‘ฅ)=๏ญ2๐‘ฅ3๐‘ฅ๐‘ฅโ‰ 0,๐‘˜๐‘ฅ=0.sincotifif

  • A6
  • B2
  • C32
  • D23

Q22:

Given ๐‘“(๐‘ฅ)=๏ฑโˆ’6๐‘ฅ๐‘ฅ<0,โˆ’6๐‘ฅ+1โˆ’๐‘ฅ๐‘ฅ๐‘ฅ>0.cosifcosif If possible or necessary, define ๐‘“(0) so that ๐‘“ is continuous at ๐‘ฅ=0.

  • AThe function cannot be made continuous at ๐‘ฅ=0 because ๐‘“(0) is undefined.
  • B๐‘“(0)=โˆ’6 makes ๐‘“ continuous at ๐‘ฅ=0.
  • CNo value of ๐‘“(0) will make ๐‘“ continuous because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ) does not exist.
  • DThe function is already continuous at ๐‘ฅ=0.

Q23:

Discuss the continuity of the function ๐‘“ at ๐‘ฅ=0 given ๐‘“(๐‘ฅ)=๏ฑ5๐‘ฅ+7๐‘ฅ2๐‘ฅ,๐‘ฅโ‰ 0,6,๐‘ฅ=0.sinsin

  • AThe function is discontinuous at ๐‘ฅ=0 because ๐‘“(0) is undefined.
  • BThe function is discontinuous at ๐‘ฅ=0 because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ)โ‰ ๐‘“(0).
  • CThe function is discontinuous at ๐‘ฅ=0 because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ) does not exist.
  • DThe function is continuous at ๐‘ฅ=0.

Q24:

Suppose ๐‘“(๐‘ฅ)=โŽงโŽจโŽฉโˆ’4๐‘ฅ+9,๐‘ฅโ‰ค3๐œ‹2,(4๐‘ฅโˆ’6๐œ‹)+13,๐‘ฅ>3๐œ‹2.sin๏Šจ What can be said of the continuity of ๐‘“ at ๐‘ฅ=3๐œ‹2?

  • AThe function is continuous at ๐‘ฅ=3๐œ‹2.
  • BThe function is discontinuous at ๐‘ฅ=3๐œ‹2 because ๐‘“๏€ผ3๐œ‹2๏ˆ is undefined.
  • CThe function is discontinuous at ๐‘ฅ=3๐œ‹2 because lim๏—โ†’๏Žข๏‘ฝ๏Žก๐‘“(๐‘ฅ) does not exist.
  • DThe function is discontinuous at ๐‘ฅ=3๐œ‹2 because ๐‘“๏€ผ3๐œ‹2๏ˆโ‰ ๐‘“(๐‘ฅ)lim๏—โ†’๏Žข๏‘ฝ๏Žก.

Q25:

Given ๐‘“(๐‘ฅ)=๏ญ4๐‘ฅ๐‘ฅ<0,โˆ’8๐‘ฅโˆ’7๐‘ฅ๐‘ฅ>0.tanifsincosif If possible or necessary, define ๐‘“(0) so that ๐‘“ is continuous at ๐‘ฅ=0.

  • A๐‘“(๐‘ฅ)=0 would make ๐‘“ continuous at ๐‘ฅ=0.
  • B๐‘“(๐‘ฅ)=โˆ’7 would make ๐‘“ continuous at ๐‘ฅ=0.
  • CThe function can not be made continuous at ๐‘ฅ=0 because lim๏—โ†’๏Šฆ๐‘“(๐‘ฅ) does not exist.
  • DThe function is already continuous at ๐‘ฅ=0.

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