Give examples of two functions f(x) and g(x) such that the following
inequality holds:
limx→0(f(x) + g(x)) != limx→0f(x) + limx→0g(x) .
Answers
Answer:
2.3 Calculating Limits Using the Limit Laws
Calculating limits by testing values of x close to a is tedious. The following Theorem essentially says
that any ‘nice’ combination of functions has exactly the limit you’d expect.
Theorem. Suppose limx→a
f(x) and limx→a
g(x) both exist and that c is constant.
Then the following limits exist and may be computed.
1. limx→a
c = c
2. limx→a
x = a
3. limx→a
c f(x) = c limx→a
f(x)
4. limx→a
[ f(x) ± g(x)] = limx→a
f(x) ± limx→a
g(x)
5. limx→a
(f(x)g(x)) = limx→a
f(x) · limx→a
g(x)
6. limx→a
f(x)
g(x)
=
limx→a
f(x)
limx→a
g(x)
Step-by-step explanation:
Examples
1. Suppose that limx→a
f(x) = 3, limx→a
g(x) = −1, lim
x→a−
h(x) = ∞, and lim
x→a+
h(x) = 6. Then
limx→a
f(x) + 3g(x) = 3 + 3(−1) = 0
lim
x→a−
2 f(x)g(x)h(x) = 2 · 3 · (−1) lim
x→a−
h(x) = −∞
lim
x→a+
2 f(x)g(x)h(x) = 2 · 3 · (−1) · 6 = 36
lim
x→a−
g(x)
f(x)h(x)
=
−1
3 lim
x→a−
h(x)
= 0
lim
x→a+
g(x)
f(x)h(x)
=
−1
3 · 6
= −
1
18
2. Simple evaluation: lim
x→1
x
3 + 2x
2 − x − 1
4x
2 − 1
=
1 + 2 − 1 − 1
4 − 1
=
1
3
3. Factorizing: lim
x→2−
x
2 − 7x + 10
x
2 − 4x + 4
= lim
x→2−
(x − 2)(x − 5)
(x − 2)(x − 2)
= lim
x→2−
x − 5
x − 2
= −3 lim
x→2−
1
x − 2
= ∞
Roots and Rationalizing
Theorem. limx→a
f(x) = L =⇒ limx→a
pn
f(x) = √n
L
Recall how you would convert an expression with surds in the denominator into one with surds
in the numerator:
4
3 +
√
5
=
4
3 +
√
5
·
3 −
√
5
3 −
√
5
=
4(3 −
√
5)
9 − 5
= 3 −
√
5
A similar approach can be used for limits.
Examples
1. lim
x→0
√
x + 3 −
√
3
x
yields the indeterminate form 0
0
. Multiplying by 1 =
√
x + 3 +
√
3
√
x + 3 +
√
3
= 1 fixes
the problem:
lim
x→0
√
x + 3 −
√
3
x
= lim
x→0
√
x + 3 −
√
3
x
·
√
x + 3 +
√
3
√
x + 3 +
√
3
= lim
x→0
x + 3 − 3
x(
√
x + 3 +
√
3)
= lim
x→0
1
√
x + 3 +
√
3
=
1
2
√
3
2. lim
x→4
√
x
2 + 9 − 5
x − 4
=
4
5
2.Comparing Limits and the Squeeze Theorem
While simple limits can be computed using the basic limit laws, more complicated functions are often
best treated by comparison.
Theorem. Suppose that f(x) ≤ g(x) for all x 6= a and suppose that limx→a
f(x) and limx→a
g(x) both exist. Then
limx→a
f(x) ≤ limx→a
g(x)
Theorem (Squeeze Theorem). Suppose that f(x) ≤ g(x) ≤ h(x) for all x 6= a, and that limx→a
f(x) =
limx→a
h(x) = L. Then limx→a
g(x) exists and also equals L.
x
f(x)
g(x)
h(x)
a
y
Example In this example we compare the complicated
function g(x) = x sin
Let the two functions f(x) and g(x) be sin(x) and cos(x) respectively.
Now,
lim x→0 (sinx + cosx) = K
⇒K = sin(0) + cos(0)
⇒K = 1
Now let
lim x→0 ( sinx + cosx) = lim x→0 sin(x) + lim x→0 cos(x)
= 0 + 1
= 1
So this proves that,
lim x→0 (f(x) + g(x)) = lim x→0 f(x) + lim x→0 g(x)
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