This document defines limits and provides examples of calculating different types of limits. It introduces: - The definition of a limit as the value a function approaches for a given input value. - Examples of calculating one-sided (right-hand and left-hand) limits and infinite limits. - Laws for calculating limits of sums, differences, products, and constant multiples. - Infinite limits as the function approaches positive even integer powers or from the left/right sides.

1. AP CALCULUS AB EXAM
LIMIT

2. WHAT IS LIMIT
It is defined as the values that a function approaches the output for the
given input values.
It is defined as the value that the function approaches as it goes to a
variable value.
Let 𝑓 𝑥 be a function defined at all values in an open interval containing
𝑎, with the possible exception of itself, and let 𝐿 be a real number. If all
values of the function 𝑓 𝑥 approaches the real number 𝐿 as the values of
𝑥 ≠ 𝑎 approach the number 𝑎, then we say that the limit of 𝑓 𝑥 as 𝑥
approaches 𝑎 is 𝐿. (In other words, as 𝑥 gets close to 𝑎, 𝑓 𝑥 gets close and
stays close to 𝐿)
lim 𝑓 𝑥 = 𝐿

3. EXAMPLE OF LIMIT
What is the limit of the function 𝑓 𝑥 = 𝑥3 as 𝑥 approaches 3 ?
lim
𝑥→3
𝑓 𝑥 = lim
𝑥→3
𝑥3
Substitute 3 for 𝑥 in the limit function.
lim
𝑥→3
𝑥3 = 33 = 27

4. RIGHT-HAND LIMIT
If 𝑥 approaches 𝑎 from the right side, i.e. from the values greater
than 𝑎, the function is said to have a right-hand limit. If 𝑞 is the
right-hand limit of 𝑓 as 𝑥 approaches 𝑎, we write as
lim
𝑥→𝑎+
𝑓 𝑥 = 𝑞

5. EXAMPLE OF RIGHT-HAND LIMIT
When 𝑥 = 3.1, 𝑓 3.1 = 29.791
When 𝑥 = 3.01, 𝑓 3.01 = 27.270901
When 𝑥 = 3.001, 𝑓 3.001 = 27.027009001
When 𝑥 = 3.0001, 𝑓 3.0001 = 27.002700090001
As 𝑥 decrease and approaches 3, 𝑓 𝑥 still approaches 27.
lim
𝑥→3+
𝑥3 = 27

6. LEFT-HAND LIMIT
If 𝑥 approaches 𝑎 from the left side, i.e. from the values lesser
than 𝑎, the function is said to have a left-hand limit. If 𝑝 is the
right-hand limit of 𝑓 as 𝑥 approaches 𝑎, we write as
lim
𝑥→𝑎−
𝑓 𝑥 = 𝑝

7. EXAMPLE OF LEFT-HAND LIMIT
When 𝑥 = 2.9, 𝑓 2.9 = 24.389
When 𝑥 = 2.99, 𝑓 2.99 = 26.730899.
When 𝑥 = 2.999, 𝑓 2.999 = 26.973008999
When 𝑥 = 2.9999, 𝑓 2.9999 = 26.997300089999
As 𝑥 increase and approaches 3, 𝑓 𝑥 still approaches 27.
lim
𝑥→3−
𝑥3 = 27

8. BASIC RULE FOR LIMIT
For any real number 𝑎 and any constant 𝑐,
lim
𝑥→𝑎
𝑥 = 𝑎
lim
𝑥→𝑎
𝑐 = 𝑎
For example:
1)
lim
𝑥→2
𝑥
Substitute 2 for 𝑥 in the limit
function.
lim
𝑥→2
𝑥 = 2
2)
lim
𝑥→2
5
The limit of a constant is that
constant.
lim
𝑥→2
5 = 5

9. SUM LAW FOR LIMIT
Let 𝑓 𝑥 and 𝑔 𝑥 be defined for all 𝑥 ≠ 𝑎 over some open interval containing 𝑎. Assume that
𝐿 and 𝑀 are real numbers such that lim
𝑥→𝑎
𝑓 𝑥 = 𝐿 and lim
𝑥→𝑎
𝑔 𝑥 = 𝑀. Let 𝑐 be a constant.
lim
𝑥→𝑎
𝑓 𝑥 + 𝑔 𝑥 = lim
𝑥→𝑎
𝑓 𝑥 + lim
𝑥→𝑎
𝑔 𝑥 = 𝐿 + 𝑀
For example:
Evaluate lim
𝑥→−3
𝑥 + 3
Use the sum law for limit, lim
𝑥→𝑎
𝑓 𝑥 + 𝑔 𝑥 = lim
𝑥→𝑎
𝑓 𝑥 + lim
𝑥→𝑎
𝑔 𝑥
lim
𝑥→−3
𝑥 + 3 = lim
𝑥→−3
𝑥+ lim
𝑥→−3
3
Use the basic rule for limit, lim
𝑥→𝑎
𝑥 = 𝑎 and lim
𝑥→𝑎
𝑐 = 𝑎
lim
𝑥→−3
𝑥+ lim
𝑥→−3
3 = −3 + 3 = 0

10. DIFFERENCE LAW FOR LIMIT
Let 𝑓 𝑥 and 𝑔 𝑥 be defined for all 𝑥 ≠ 𝑎 over some open interval containing 𝑎. Assume that 𝐿 and 𝑀 are
real numbers such that lim
𝑥→𝑎
𝑓 𝑥 = 𝐿 and lim
𝑥→𝑎
𝑔 𝑥 = 𝑀. Let 𝑐 be a constant.
lim
𝑥→𝑎
𝑓 𝑥 − 𝑔 𝑥 = lim
𝑥→𝑎
𝑓 𝑥 − lim
𝑥→𝑎
𝑔 𝑥 = 𝐿 − 𝑀
For example:
Evaluate lim
𝑥→3
𝑥 − 5
Use the difference law for limit, lim
𝑥→𝑎
𝑓 𝑥 − 𝑔 𝑥 = lim
𝑥→𝑎
𝑓 𝑥 − lim
𝑥→𝑎
𝑔 𝑥
lim
𝑥→3
𝑥 − 3 = lim
𝑥→3
𝑥 − lim
𝑥→3
5
Use the basic rule for limit, lim
𝑥→𝑎
𝑥 = 𝑎 and lim
𝑥→𝑎
𝑐 = 𝑎
lim
𝑥→3
𝑥 − lim
𝑥→3
5 = 3 − 5 = −2

11. PRODUCT LAW FOR LIMIT
Let 𝑓 𝑥 and 𝑔 𝑥 be defined for all 𝑥 ≠ 𝑎 over some open interval containing 𝑎. Assume that
𝐿 and 𝑀 are real numbers such that lim
𝑥→𝑎
𝑓 𝑥 = 𝐿 and lim
𝑥→𝑎
𝑔 𝑥 = 𝑀. Let 𝑐 be a constant.
lim
𝑥→𝑎
𝑓 𝑥 ∙ 𝑔 𝑥 = lim
𝑥→𝑎
𝑓 𝑥 ∙ lim
𝑥→𝑎
𝑔 𝑥 = 𝐿 ∙ 𝑀
For example:
Evaluate lim
𝑥→3
𝑥 𝑥 + 5
Use the product law for limit, lim
𝑥→𝑎
𝑓 𝑥 ∙ 𝑔 𝑥 = lim
𝑥→𝑎
𝑓 𝑥 ∙ lim
𝑥→𝑎
𝑔 𝑥
lim
𝑥→3
𝑥 𝑥 + 5 = lim
𝑥→3
𝑥 ∙ lim
𝑥→3
𝑥 + 5
Use the sum law for limit, lim
𝑥→𝑎
𝑝 𝑥 + 𝑞 𝑥 = lim
𝑥→𝑎
𝑝 𝑥 + lim
𝑥→𝑎
𝑞 𝑥
lim
𝑥→3
𝑥 ∙ lim
𝑥→3
𝑥 + 5 = lim
𝑥→3
𝑥 ∙ lim
𝑥→3
𝑥 + lim
𝑥→3
5
Use the basic rule for limit, lim
𝑥→𝑎
𝑥 = 𝑎 and lim
𝑥→𝑎
𝑐 = 𝑎
lim
𝑥→3
𝑥 ∙ lim
𝑥→3
𝑥 + lim
𝑥→3
5 = 3 3 + 5 = 3 8 = 24

12. CONSTANT MULTIPLE LAW FOR LIMIT
Let 𝑓 𝑥 be defined for all 𝑥 ≠ 𝑎 over some open interval containing 𝑎. Assume that 𝐿 are real numbers
such that lim
𝑥→𝑎
𝑓 𝑥 = 𝐿. Let 𝑐 be a constant.
lim
𝑥→𝑎
𝑐𝑓 𝑥 = 𝑐 lim
𝑥→𝑎
𝑓 𝑥 = 𝑐𝐿
For example:
Evaluate lim
𝑥→3
5𝑥2
.
Use the constant multiple law for limit, lim
𝑥→𝑎
𝑐𝑓 𝑥 = 𝑐 lim
𝑥→𝑎
𝑓 𝑥
lim
𝑥→3
5𝑥2
= 5 lim
𝑥→3
𝑥2
Substitute 5 for 𝑥 in 5 lim
𝑥→3
𝑥2
.
5 lim
𝑥→3
𝑥2
= 5 32
= 5 9 = 45
Type equation here.

13. INFINITE LIMIT FROM THE LEFT
Let 𝑓 𝑥 be a function defined at all values in an open interval of the form
𝑏, 𝑎 .
If the values of 𝑓 𝑥 increase without bound as the values of 𝑥 (where 𝑥 < 𝑎)
approach the number 𝑎, then we sat that the limit as 𝑥 approaches 𝑎 from the
left is positive infinity.
lim
𝑥→𝑎−
𝑓 𝑥 = −∞
If the values of 𝑓 𝑥 decrease without bound as the values of 𝑥 (where 𝑥 < 𝑎)
approach the number 𝑎, then we sat that the limit as 𝑥 approaches 𝑎 from the
left is negative infinity.
lim
𝑥→𝑎−
𝑓 𝑥 = +∞

14. EXAMPLE OF CASE 1
Evaluate the limit lim
𝑥→0−
−
1
𝑥
, if possible.
Here, 𝑓 𝑥 = −
1
𝑥
When 𝑥 = −0.1, 𝑓 0.1 = 10
When 𝑥 = −0.01, 𝑓 0.01 = 100
When 𝑥 = −0.001, 𝑓 0.001 = 1000
When 𝑥 = −0.0001, 𝑓 0.0001 = 10,000
The value of 𝑓 𝑥 increase without bound as 𝑥 approaches 0 from the left.
lim
𝑥→0−
−
1
𝑥
= +∞

15. EXAMPLE OF CASE 2
Evaluate the limit lim
𝑥→0−
1
𝑥
, if possible.
Here, 𝑓 𝑥 =
1
𝑥
When 𝑥 = −0.1, 𝑓 0.1 = −10
When 𝑥 = −0.01, 𝑓 0.01 = −100
When 𝑥 = −0.001, 𝑓 0.001 = −1000
When 𝑥 = −0.0001, 𝑓 0.0001 = −10,000
The value of 𝑓 𝑥 decrease without bound as 𝑥 approaches 0 from the left.
lim
𝑥→0−
1
𝑥
= −∞

16. INFINITE LIMIT FROM THE RIGHT
Let 𝑓 𝑥 be a function defined at all values in an open interval of the form
𝑎, 𝑐 .
If the values of 𝑓 𝑥 increase without bound as the values of 𝑥 (where 𝑥 > 𝑎)
approach the number 𝑎, then we sat that the limit as 𝑥 approaches 𝑎 from the
right is positive infinity.
lim
𝑥→𝑎+
𝑓 𝑥 = +∞
If the values of 𝑓 𝑥 decrease without bound as the values of 𝑥 (where 𝑥 > 𝑎)
approach the number 𝑎, then we sat that the limit as 𝑥 approaches 𝑎 from the
right is negative infinity.
lim
𝑥→𝑎+
𝑓 𝑥 = −∞

17. EXAMPLE OF CASE 1
Evaluate the limit lim
𝑥→0+
1
𝑥
, if possible.
Here, 𝑓 𝑥 =
1
𝑥
When 𝑥 = 0.1, 𝑓 0.1 = 10
When 𝑥 = 0.01, 𝑓 0.01 = 100
When 𝑥 = 0.001, 𝑓 0.001 = 1000
When 𝑥 = 0.0001, 𝑓 0.0001 = 10,000
The value of 𝑓 𝑥 increase without bound as 𝑥 approaches 0 from the right.
lim
𝑥→0+
1
𝑥
= +∞

18. EXAMPLE FOR CASE 2
Evaluate the limit lim
𝑥→0+
−
1
𝑥
, if possible.
Here, 𝑓 𝑥 = −
1
𝑥
When 𝑥 = 0.1, 𝑓 0.1 = −10
When 𝑥 = 0.01, 𝑓 0.01 = −100
When 𝑥 = 0.001, 𝑓 0.001 = −1000
When 𝑥 = 0.0001, 𝑓 0.0001 = −10,000
The value of 𝑓 𝑥 decrease without bound as 𝑥 approaches 0 from the right.
lim
𝑥→0+
−
1
𝑥
= −∞

19. INFINITE LIMITS FROM POSITIVE EVEN
INTEGERS
If 𝑛 is a positive even integer, then
lim
𝑥→𝑎
1
𝑥 − 𝑎 𝑛
= +∞
lim
𝑥→𝑎+
1
𝑥 − 𝑎 𝑛
= +∞
lim
𝑥→𝑎−
1
𝑥 − 𝑎 𝑛
= +∞

20. EXAMPLE
Evaluate lim
𝑥→−3
1
𝑥+3 4
lim
𝑥→−3
1
(−3) + 3 4 =
1
0
= +∞
Evaluate lim
𝑥→−3+
1
𝑥+3 4
lim
ℎ→0
1
(−3 + ℎ) + 3 4
= lim
ℎ→0
1
ℎ4
=
1
0
= +∞
Evaluate lim
𝑥→−3−
1
𝑥+3 4
lim
ℎ→0
1
(−3 − ℎ) + 3 4
= lim
ℎ→0
1
(−ℎ)4
= lim
ℎ→0
1
ℎ4
=
1
0
= +∞

21. INFINITE LIMITS FROM POSITIVE ODD
INTEGERS
If 𝑛 is a positive odd integer, then
lim
𝑥→𝑎
1
𝑥 − 𝑎 𝑛
= 𝐷𝑁𝐸
lim
𝑥→𝑎+
1
𝑥 − 𝑎 𝑛
= +∞
lim
𝑥→𝑎−
1
𝑥 − 𝑎 𝑛
= −∞

22. EXAMPLE
Evaluate lim
𝑥→−3+
1
𝑥+3 3
lim
ℎ→0
1
(−3 + ℎ) + 3 3
= lim
ℎ→0
1
ℎ3
=
1
0
= +∞
Evaluate lim
𝑥→−3−
1
𝑥+3 3
lim
ℎ→0
1
(−3 − ℎ) + 3 3
= lim
ℎ→0
1
(−ℎ)3
= −lim
ℎ→0
1
ℎ3
= −
1
0
= −∞
Evaluate lim
𝑥→−3
1
𝑥+3 3
lim
𝑥→−3
1
(−3) + 3 3
= 𝐷𝑁𝐸
Because lim
𝑥→−3+
1
𝑥+3 3 ≠ lim
𝑥→−3−
1
𝑥+3 3

23. LIMIT AT INFINITY FOR RATIONAL
FUNCTION
For rational function 𝑓 𝑥 =
𝑝 𝑥
𝑞 𝑥
, the limit at infinity is determined by
the relationship between the degree of 𝑝 and 𝑞.
If the degree of 𝑝 is less than the degree of 𝑞, then the line 𝑦 = 0 is a
horizontal asymptote for 𝑓.
If the degree of 𝑝 is equal to the degree of 𝑞, then the line 𝑦 =
𝑎𝑛
𝑏𝑛
is a
horizontal asymptote for 𝑓, where 𝑎𝑛 and 𝑏𝑛 are the leading
coefficients of 𝑝 and 𝑞.
If the degree of 𝑝 is greater than the degree of 𝑞, then 𝑓 approaches ∞
or −∞ at each end.

24. EXAMPLE FOR CASE 1
Evaluate lim
𝑥→∞
3𝑥2+2𝑥
4𝑥3−5𝑥+7
lim
𝑥→∞
3𝑥2+2𝑥
4𝑥3−5𝑥+7
= lim
𝑥→∞
3𝑥2+2𝑥
𝑥3 4−
5
𝑥2+
7
𝑥3
= lim
𝑥→∞
3
𝑥
+
2
𝑥2
4−
5
𝑥2+
7
𝑥3
=
3(0) + 2(0)
4 − 5(0) + 7(0)
=
0
4
= 0

25. EXAMPLE FOR CASE 2
Evaluate lim
𝑥→∞
2𝑥+3
3𝑥−2
lim
𝑥→∞
2𝑥+3
3𝑥−2
= lim
𝑥→∞
𝑥 2+
3
𝑥
𝑥 3−
2
𝑥
= lim
𝑥→∞
2+
3
𝑥
3−
2
𝑥
=
2 + 0
3 − 0
=
2
3

26. EXAMPLE FOR CASE 3
Evaluate lim
𝑥→∞
3𝑥2+4𝑥
𝑥+2
lim
𝑥→∞
3𝑥2+4𝑥
𝑥+2
= lim
𝑥→∞
𝑥2 3+
4
𝑥
𝑥 1+
2
𝑥
= lim
𝑥→∞
𝑥 3+
4
𝑥
1+
2
𝑥
=
∞ 3 + 0
1 + 0
=
∞
1
=∞