Limits
Anibal Garcia
May 8, 2020
Contents
1 Indeterminate Form k0 , k 6= 0 2
0
2 Indeterminate Form 0
4
∞
3 Indeterminate Form ∞
7
4 Indeterminate Form ∞ − ∞ 12
5 Indeterminate Form 0 · ∞ 16
1
,1 Indeterminate Form k0 , k 6= 0
If we get an indeterminate form of k0 after calculating the limit of a function, we
will need to calculate the lateral limits.
If a the function has a limit, both laterals will have the same value and therefore
the limit will exist. If the values do not match, the limit for the function will not
exist.
Example 1:
3x + 7
lim
x→0 2x4
After substituting x for 0, we verify that:
3x + 7 7
lim 4
−→
x→0 2x 0
Therefore, we find the lateral limits:
3x + 7
lim+ = +∞
x→0 2x4
3x + 7
lim− = +∞
x→0 2x4
Both lateral limits have the same value ∴
3x + 7 7
∃ lim 4
→ = +∞
x→0 2x 0
2
, Example 2:
We calculate the limit for the function:
6x + 2
lim1
x→ 2 4x − 2
1
After substituting x for 2
we verify that:
6x + 2 5
lim1 −→
x→ 2 4x − 2 0
We then find the values for the lateral limits:
6x + 2 5
lim+ = + = +∞
x→ 12 4x − 2 0
6x + 2 5
lim− = − = −∞
x→ 12 4x − 2 0
The values for the lateral limits don’t match:
6x + 2
∴ @ lim1
x→ 2 4x − 2
3
Anibal Garcia
May 8, 2020
Contents
1 Indeterminate Form k0 , k 6= 0 2
0
2 Indeterminate Form 0
4
∞
3 Indeterminate Form ∞
7
4 Indeterminate Form ∞ − ∞ 12
5 Indeterminate Form 0 · ∞ 16
1
,1 Indeterminate Form k0 , k 6= 0
If we get an indeterminate form of k0 after calculating the limit of a function, we
will need to calculate the lateral limits.
If a the function has a limit, both laterals will have the same value and therefore
the limit will exist. If the values do not match, the limit for the function will not
exist.
Example 1:
3x + 7
lim
x→0 2x4
After substituting x for 0, we verify that:
3x + 7 7
lim 4
−→
x→0 2x 0
Therefore, we find the lateral limits:
3x + 7
lim+ = +∞
x→0 2x4
3x + 7
lim− = +∞
x→0 2x4
Both lateral limits have the same value ∴
3x + 7 7
∃ lim 4
→ = +∞
x→0 2x 0
2
, Example 2:
We calculate the limit for the function:
6x + 2
lim1
x→ 2 4x − 2
1
After substituting x for 2
we verify that:
6x + 2 5
lim1 −→
x→ 2 4x − 2 0
We then find the values for the lateral limits:
6x + 2 5
lim+ = + = +∞
x→ 12 4x − 2 0
6x + 2 5
lim− = − = −∞
x→ 12 4x − 2 0
The values for the lateral limits don’t match:
6x + 2
∴ @ lim1
x→ 2 4x − 2
3
