Calculus 1
LIMITS
Instruction: Evaluate the following limits. Show your solution.
PROBLEMS
(1. ) lim (
𝑥3
+ 3𝑥 + 5
)
− 6𝑥 + 1
𝑥→∞ 2𝑥3
𝑥 20 − 1
(2. ) lim ( 10
)
𝑥→1 𝑥
−1
SOLUTIONS
3
5
𝑥 3 (1 + 2 + 3 )
𝑥 3 + 3𝑥 + 5
𝑥
𝑥
lim (
) = lim (
)
6
1
𝑥→∞ 2𝑥 3 − 6𝑥 + 1
𝑥→∞
𝑥 3 (2 − 2 + 3 )
𝑥
𝑥
3
5
(1 + 2 + 3 )
𝑥
𝑥 )
= lim (
6
1
𝑥→∞
(2 − 2 + 3 )
𝑥
𝑥
(1 + 0 + 0)
=
(2 − 0 + 0)
𝒙𝟑 + 𝟑𝒙 + 𝟓
𝟏
𝐥𝐢𝐦 ( 𝟑
)=
𝒙→∞ 𝟐𝒙 − 𝟔𝒙 + 𝟏
𝟐
lim (
𝑥→1
(𝑥 10 − 1)(𝑥 10 + 1)
𝑥 20 − 1
)
=
lim
(
)
𝑥→1
(𝑥 10 − 1)
𝑥 10 − 1
= lim (𝑥 10 + 1)
𝑥→1
= (1)10 + 1
=1+1
𝒙𝟐𝟎 − 𝟏
𝐥𝐢𝐦 ( 𝟏𝟎
)=𝟐
𝒙→𝟏 𝒙
−𝟏
(3. ) lim (
𝑦→−2
𝑦 3 + 3𝑦 2 + 2𝑦
)
𝑦2 − 𝑦 − 6
lim (
𝑦→−2
(𝑦)(𝑦 + 2)(𝑦 + 1)
𝑦 3 + 3𝑦 2 + 2𝑦
) = lim (
)
2
𝑦→−2
(𝑦 + 2)(𝑦 − 3)
𝑦 −𝑦−6
(𝑦)(𝑦 + 1)
= lim (
)
𝑦→−2
(𝑦 − 3)
𝑦2 + 𝑦
= lim (
)
𝑦→−2 𝑦 − 3
(−2)2 + (−2)
=
(−2) − 3
4−2
=
−5
2
=
−5
𝐥𝐢𝐦 (
𝒚→−𝟐
3
3
𝒚𝟑 + 𝟑𝒚𝟐 + 𝟐𝒚
𝟐
)=−
𝟐
𝒚 −𝒚−𝟔
𝟓
3
3
√𝑥 − 1
(4. ) lim (
)
𝑥→1
𝑥−1
3
√𝑥 − 1
√𝑥 − 1 ( √𝑥 2 + √𝑥 + 1)
lim (
) = lim (
× 3
)
3
𝑥→1
𝑥→1
𝑥−1
𝑥−1
( √𝑥 2 + √𝑥 + 1)
(𝑥 − 1)
= lim (
)
𝑥→1 (𝑥 − 1)( 3√𝑥 2 + 3√𝑥 + 1)
1
= lim ( 3
)
2
𝑥→1 ( √𝑥 + 3√𝑥 + 1)
1
= 3
( √(1)2 + 3√(1) + 1)
1
=
(1 + 1 + 1)
𝟑
𝟏
√𝒙 − 𝟏
𝐥𝐢𝐦 (
)=
𝒙→𝟏
𝒙−𝟏
𝟑
𝑦 2 − 4𝑦 − 21
(5. ) lim ( 2
)
𝑦→7 3𝑦 − 17𝑦 − 28
(𝑦 + 3)(𝑦 − 7)
𝑦 2 − 4𝑦 − 21
lim ( 2
) = lim (
)
𝑦→7 3𝑦 − 17𝑦 − 28
𝑦→7 (3𝑦 + 4)(𝑦 − 7)
(𝑦 + 3)
= lim (
)
𝑦→7 (3𝑦 + 4)
(7) + 3
=
3(7) + 4
10
=
21 + 4
10
=
25
𝒚𝟐 − 𝟒𝒚 − 𝟐𝟏
𝟐
𝐥𝐢𝐦 ( 𝟐
)=
𝒚→𝟕 𝟑𝒚 − 𝟏𝟕𝒚 − 𝟐𝟖
𝟓