Worksheet: Existence of Limits

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In this worksheet, we will practice determining whether the limit of a function at a certain value exists.

Q1:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=−15𝑥−15𝑥<−15,𝑥−15𝑥≥−15.ifif

  • AThe limit does not exist because lim→𝑓(𝑥) exists, but lim→𝑓(𝑥) does not exist.
  • BThe limit does not exist because both lim→𝑓(𝑥) and lim→𝑓(𝑥) exist, but are not equal.
  • CThe limit exists and equals 210.
  • DThe limit does not exist because lim→𝑓(𝑥) exists, but lim→𝑓(𝑥) does not exist.
  • EThe limit exists and equals −15.

Q2:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=𝑥−4𝑥<−1,20𝑥>−1.ifif

  • AThe limit does not exist because 𝑓(−1)≠𝑓(−1).
  • BThe limit does not exist because 𝑓(−1) does not exist.
  • CThe limit exists and equals 20.
  • DThe limit exists and equals −20.

Q3:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=⎧⎪⎨⎪⎩2622−𝑥𝑥<4,√𝑥−2𝑥−4𝑥>4.ifif

  • AThe limit does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).
  • BThe limit exists and equals 5.
  • CThe limit does not exist because lim→𝑓(𝑥) does not exist.
  • DThe limit exists and equals 139.
  • EThe limit does not exist because lim→𝑓(𝑥) does not exist.

Q4:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=𝑥|𝑥+7|−3.

  • AThe limit does not exist because lim→𝑓(𝑥) does not exist.
  • BThe limit exists and equals −3.
  • CThe limit exists and equals −9.
  • DThe limit does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).
  • EThe limit does not exist because lim→𝑓(𝑥) does not exist.

Q5:

Given that 𝑓(𝑥)=4+𝑥+3𝑥|𝑥+3|−3<𝑥<0,2𝑥+40<𝑥<2,ififfind lim→𝑓(𝑥).

Q6:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=16𝑥+|𝑥−2|2−𝑥𝑥<2,𝑥+5𝑥>2.ifif

  • AThe limit exists and equals 33.
  • BThe limit exists and equals −22.
  • CThe limit does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).
  • DThe limit does not exist because lim→𝑓(𝑥) does not exist.
  • EThe limit does not exist because lim→𝑓(𝑥) does not exist.

Q7:

Find lim→𝑓(𝑥) given 𝑓(𝑥)=5𝑥+3𝑥<−1,2𝑥−1<𝑥<5,𝑥+4𝑥>5.ififif

  • A10
  • B28
  • C − 2

Q8:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=⎧⎪⎪⎨⎪⎪⎩3𝑥−𝜋2<𝑥<0,3𝑥+10<𝑥<3,𝑥−27𝑥−3𝑥>3.cosififif

  • A l i m → 𝑓 ( 𝑥 ) does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).
  • B l i m → 𝑓 ( 𝑥 ) exist and equals 3.
  • C l i m → 𝑓 ( 𝑥 ) does not exist because lim→𝑓(𝑥) exist, but lim→𝑓(𝑥) does not exist.
  • D l i m → 𝑓 ( 𝑥 ) does not exist because lim→𝑓(𝑥) exist, but lim→𝑓(𝑥) does not exist.

Q9:

Given that the function 𝑓(𝑥)=𝑥+𝑎𝑥+𝑏𝑥−5𝑥+6𝑥<2,6𝑥𝑥>2ifif has a limit when 𝑥=2, determine the values of 𝑎 and 𝑏.

  • A 𝑎 = − 1 2 , 𝑏 = 2 8
  • B 𝑎 = − 1 6 , 𝑏 = 3 6
  • C 𝑎 = − 1 6 , 𝑏 = 2 8
  • D 𝑎 = − 8 , 𝑏 = − 1 2

Q10:

Determine lim→𝑓(𝑥) using the graph.

  • AThe limit does not exist.
  • B4
  • C − 7
  • D − 5

Q11:

Given that 𝑓(𝑥)=⎧⎨⎩𝑥+22𝑥<3,5𝑥−10𝑥−152𝑥−4𝑥−6𝑥>3,ififdetermine lim→𝑓(𝑥).

  • A − 5 2
  • B 5 2
  • C 5 4
  • D5

Q12:

Given that 𝑓(𝑥)=𝑥+7√𝑥𝑥, determine lim→𝑓(𝑥).

  • A7
  • B0
  • CThe limit does not exist.
  • D − 7

Q13:

Given the function 𝑓(𝑥)=2𝑥|𝑥−4|+5, what is true of lim→𝑓(𝑥)?

  • AThe limit exists and equals 13.
  • BThe limit does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).
  • CThe limit exists and equals −3.

Q14:

Determine lim→𝑓(𝑥).

Q15:

What can be said of lim→𝑓(𝑥) for the function 𝑓(𝑥)=7𝑥|𝑥|+9𝑥<0,4|𝑥|𝑥+5𝑥>0?ifif

  • A l i m → 𝑓 ( 𝑥 ) exists and equals 9.
  • B l i m → 𝑓 ( 𝑥 ) exists and equals −1.
  • C l i m → 𝑓 ( 𝑥 ) does not exist because lim→𝑓(𝑥) exists, but lim→𝑓(𝑥) does not exist.
  • D l i m → 𝑓 ( 𝑥 ) exists and equals −9.
  • E l i m → 𝑓 ( 𝑥 ) does not exist because lim→𝑓(𝑥) exists, but lim→𝑓(𝑥) does not exist.

Q16:

Given that 𝑓(𝑥)=4𝑥+2𝑥<1,𝑥+3𝑥+61<𝑥<5,4𝑥+30𝑥>5,ififif find lim→𝑓(𝑥), if it exists.

Q17:

Find the values of 𝑎 and 𝑏 given lim→𝑓(𝑥) and lim→𝑓(𝑥) exist where 𝑓(𝑥)=7𝑥−9𝑥<−1,𝑎𝑥+𝑏−1<𝑥<5,−8𝑥−6𝑥>5.ififif

  • A 𝑎 = − 5 , 𝑏 = − 2 1
  • B 𝑎 = − 3 7 , 𝑏 = − 5 3
  • C 𝑎 = 3 7 , 𝑏 = 2 1
  • D 𝑎 = 5 , 𝑏 = − 1 1
  • E 𝑎 = 2 7 , 𝑏 = 1 1

Q18:

Suppose that 𝑓(𝑥)=⎧⎪⎨⎪⎩√𝑥−1𝑥−1𝑥<1,√𝑥−18√𝑥+4√𝑥+4𝑥−1𝑥>1.ifif What can be said about the existence of lim→𝑓(𝑥)?

  • AThe limit exists and equals 67.
  • BThe limit exists and equals 85.
  • CThe limit exists and equals 0.
  • DThe limit does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).

Q19:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=|𝑥−2|+3−2<𝑥<3,𝑥+6𝑥−27𝑥−3𝑥3<𝑥<9.ifif

  • A l i m → 𝑓 ( 𝑥 ) exists and equals 4.
  • B l i m → 𝑓 ( 𝑥 ) does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).
  • C l i m → 𝑓 ( 𝑥 ) does not exist because lim→𝑓(𝑥) exist, but lim→𝑓(𝑥) does not exist.
  • D l i m → 𝑓 ( 𝑥 ) does not exist because lim→𝑓(𝑥) exist, but lim→𝑓(𝑥) does not exist.

Q20:

The figure shows the graph of the function 𝑓(𝑥)=2𝜋𝑥𝑥>0,0.5𝑥≤0.sin

What is the 𝑥-coordinate of point 𝐴? Give your answer as a fraction.

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

What is the 𝑥-coordinate of point 𝐵? Give your answer as a fraction.

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

Setting the 𝑥-coordinate of 𝐴 above as 𝑎, give a formula of the sequence 𝑎, 𝑛≥1, with 𝐵=(𝑎,0.5).

  • A 𝑎 = 1 2 1 2 𝑛 + 1
  • B 𝑎 = − 1 4 4 3 2 5 𝑛 + 4 4 4 3 2 5
  • C 𝑎 = − 1 4 4 3 2 5 𝑛 + 1 2 1 3
  • D 𝑎 = 1 2 1 2 𝑛 + 1
  • E 𝑎 = 1 2 1 2 𝑛 + 1 3

What does evaluation of 𝑓(𝑎) with 𝑎 that was previously obtained tell you about a possible value for lim→𝑓(𝑥)?

What does evaluation of 𝑓(𝑎) with 𝑎=1𝑛 tell you about a possible value for lim→𝑓(𝑥)?

What, therefore, can we conclude about lim→𝑓(𝑥)?

  • AWe cannot conclude anything.
  • B l i m → 𝑓 ( 𝑥 ) = − 1
  • CWe conclude that lim→𝑓(𝑥)=0.5.
  • D l i m → 𝑓 ( 𝑥 ) = 1
  • EWe conclude that the one-sided limit does not exist.

What law of limits is used to prove that lim→𝑓(𝑥)=0.5?

  • AThe limit of the quotient of functions is the quotient of the limits of the given functions if they exist and the denominator is nonzero.
  • BThe limit of a difference of functions is the difference of the limits of the given functions if they exist.
  • CThe limit of the product of functions is the product of the limits of the given functions if they exist.
  • DThe limit of a sum of functions is the sum of the limits of the summand functions if they exist.
  • EThe limit of a constant function as 𝑥 approaches any point is the constant itself.

What can be said about lim→𝑓(𝑥)?

  • AIt does not exist because lim→𝑓(𝑥) does not exist.
  • BIt exists for some points on the number line and does not for other points.
  • CIt exists and is sometimes 0.5 and sometimes 0.
  • DIt exists and is equal to 0.5 because this is lim→𝑓(𝑥).
  • EIt exists and is any value that you like.

Q21:

Discuss the existence of lim→[(𝑓+𝑔)(𝑥)], where 𝑓(𝑥)=𝑥+8𝑥<1,𝑥−6𝑥≥1ifif and 𝑔(𝑥)=−6𝑥+𝑥𝑥<1,9𝑥𝑥≥1.ifif

  • A l i m → [ ( 𝑓 + 𝑔 ) ( 𝑥 ) ] exists and equals 9.
  • B l i m → [ ( 𝑓 + 𝑔 ) ( 𝑥 ) ] exists and equals −5.
  • C l i m → [ ( 𝑓 + 𝑔 ) ( 𝑥 ) ] does not exist because lim→[(𝑓+𝑔)(𝑥)] exists, but lim→[(𝑓+𝑔)(𝑥)] does not exist.
  • D l i m → [ ( 𝑓 + 𝑔 ) ( 𝑥 ) ] does not exist because lim→[(𝑓+𝑔)(𝑥)] exists, but lim→[(𝑓+𝑔)(𝑥)] does not exist.
  • E l i m → [ ( 𝑓 + 𝑔 ) ( 𝑥 ) ] exists and equals 4.

Q22:

Determine lim→𝑓(𝑥).

  • AThe limit does not exist.
  • B10
  • C3
  • D0

Q23:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=20−2𝑥+𝑥|𝑥|−6<𝑥<2,5(𝑥−2)√𝑥+2−22<𝑥<14.ifif

  • AThe limit does not exist becauselimlim→→𝑓(𝑥)≠𝑓(𝑥).
  • BThe limit exists and equals 24.
  • CThe limit exists and equals 5.
  • DThe limit exists and equals 20.
  • EThe limit exists and equals 0.

Q24:

Given that 𝑓(𝑥)=⎧⎪⎨⎪⎩8𝑥−12𝑥+5𝑥−3𝑥<12,4𝑥−1𝑥>12,ififdetermine lim→𝑓(𝑥).

  • A1
  • BThe limit does not exist.
  • C0
  • D 6 7

Q25:

Discuss the existence of lim→𝑓(𝑥) given 𝑓(𝑥)=⎧⎨⎩9𝑥−|6𝑥−3|−6𝑥+3𝑥<12,|3𝑥+2|𝑥>12.ifif

  • AThe limit exists and equals −72.
  • BThe limit exists and equals 72.
  • CThe limit exists and equals −112.
  • DThe limit exists and equals 112.
  • EThe limit does not exist because limlim→→𝑓(𝑥)≠𝑓(𝑥).

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