3. Limit Laws
Suppose that c is a constant and the limits
lim f (x) and lim g (x)
x→a x→a
exist. Then
1. lim [f (x) + g (x)] = lim f (x) + lim g (x)
x→a x→a x→a
2. lim [f (x) − g (x)] = lim f (x) − lim g (x)
x→a x→a x→a
3. lim [cf (x)] = c lim f (x)
x→a x→a
4. lim [f (x)g (x)] = lim f (x) · lim g (x)
x→a x→a x→a
4.
5. Limit Laws, continued
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
6. Limit Laws, continued
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x)
x→a x→a
7.
8. Limit Laws, continued
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x) (follows from 3 repeatedly)
x→a x→a
9. Limit Laws, continued
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x) (follows from 3 repeatedly)
x→a x→a
7. lim c = c
x→a
8. lim x = a
x→a
10.
11.
12.
13.
14.
15. Limit Laws, continued
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x) (follows from 3 repeatedly)
x→a x→a
7. lim c = c
x→a
8. lim x = a
x→a
9. lim x n = an
x→a
√ √
10. lim n x = n a
x→a
16. Limit Laws, continued
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x) (follows from 3 repeatedly)
x→a x→a
7. lim c = c
x→a
8. lim x = a
x→a
9. lim x n = an (follows from 6 and 8)
x→a
√ √
10. lim n x = n a
x→a
17. Limit Laws, continued
lim f (x)
f (x)
= x→a
5. lim , if lim g (x) = 0.
x→a g (x) lim g (x) x→a
x→a
n
n
6. lim [f (x)] = lim f (x) (follows from 3 repeatedly)
x→a x→a
7. lim c = c
x→a
8. lim x = a
x→a
9. lim x n = an (follows from 6 and 8)
x→a
√ √
10. lim n x = n a
x→a
n
11. lim f (x) = lim f (x) (If n is even, we must additionally
n
x→a x→a
assume that lim f (x) > 0)
x→a
18. Direct Substitution Property
Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of
f , then
lim f (x) = f (a)
x→a
19. Limits do not see the point! (in a good way)
Theorem
If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
x→a x→a
20.
21. Limits do not see the point! (in a good way)
Theorem
If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
x→a x→a
Example
x 2 + 2x + 1
Find lim , if it exists.
x +1
x→−1
22. Limits do not see the point! (in a good way)
Theorem
If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
x→a x→a
Example
x 2 + 2x + 1
Find lim , if it exists.
x +1
x→−1
Solution
x 2 + 2x + 1
= x + 1 whenever x = −1, and since
Since
x +1
x 2 + 2x + 1
lim x + 1 = 0, we have lim = 0.
x +1
x→−1 x→−1
23. Finding limits by algebraic manipulations
Example √
x −2
Find lim .
x −4
x→4
24. Finding limits by algebraic manipulations
Example √
x −2
Find lim .
x −4
x→4
Solution √ √ √
2
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
25. Finding limits by algebraic manipulations
Example √
x −2
Find lim .
x −4
x→4
Solution √2 √ √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
So
√ √
x −2 x −2
= lim √ √
lim
x→2 x − 4 x→2 ( x − 2)( x + 2)
1 1
= lim √ =
4
x +2
x→2
26. Finding limits by algebraic manipulations
Example √
x −2
Find lim .
x −4
x→4
Solution √2 √ √
Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
So
√ √
x −2 x −2
= lim √ √
lim
x→2 x − 4 x→2 ( x − 2)( x + 2)
1 1
= lim √ =
4
x +2
x→2
Example√ √
x− 3a
3
Try lim .
x −a
x→a
27. Two More Important Limit Theorems
Theorem
If f (x) ≤ g (x) when x is near a (except possibly at a), then
lim f (x) ≤ lim g (x)
x→a x→a
(as usual, provided these limits exist).
Theorem (The Squeeze/Sandwich/Pinching Theorem)
If f (x) ≤ g (x) ≤ h(x) when x is near a (as usual, except possibly
at a), and
lim f (x) = lim h(x) = L,
x→a x→a
then
lim g (x) = L.
x→a
28.
29. We can use the Squeeze Theorem to make complicated limits
simple.
30. We can use the Squeeze Theorem to make complicated limits
simple.
Example
1
Show that lim x 2 sin = 0.
x
x→0
31.
32. We can use the Squeeze Theorem to make complicated limits
simple.
Example
1
Show that lim x 2 sin = 0.
x
x→0
Solution
We have for all x,
1
−x 2 ≤ x 2 sin ≤ x2
x
The left and right sides go to zero as x → 0.
