Math 121 - Summary of Limit Laws
G.Pugh
May 7 2012
G.Pugh (VIU)
Math 121 - Summary of Limit Laws
May 7 2012
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Limit Laws
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Math 121 - Summary of Limit Laws
May 7 2012
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Assumptions
In the following, suppose:
c represents a constant (a fixed number)
The limits
lim f (x)
x→a
and
lim g(x)
x→a
both exist
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Math 121 - Summary of Limit Laws
May 7 2012
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Sum Law
lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
x→a
In words: The limit of a sum is the sum of the limits
Example: lim
x→π
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√
√ x + sin x = lim x + lim sin x
x→π
Math 121 - Summary of Limit Laws
x→π
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Difference Law
lim [f (x) − g(x)] = lim f (x) − lim g(x)
x→a
x→a
x→a
In words: The limit of a difference is the difference of the limits
Example: lim
x→−3
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1
1
3
3
−x =
lim
− lim x
x
x→−3 x
x→−3
Math 121 - Summary of Limit Laws
May 7 2012
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Constant Multiplier Law
lim [cf (x)] = c lim f (x)
x→a
x→a
In words: The limit of a constant times a function is the constant
times the limit of the function.
Example: lim
√
x→ 2
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3
√
7 x
3
=
7
1
lim
√ √
x
x→ 2
Math 121 - Summary of Limit Laws
May 7 2012
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Product Law
lim [f (x)g(x)] =
x→a
lim f (x)
x→a
lim g(x)
x→a
In words: The limit of a product is the product of the limits
Example:
h
i
2
2
lim (x + 2)(1 + cos x) = lim (x + 2)
lim (1 + cos x)
x→0
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x→0
Math 121 - Summary of Limit Laws
x→0
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Quotient Law
lim f (x)
f (x)
provided lim g(x) 6= 0 .
lim
= x→a
x→a
x→a g(x)
lim g(x)
x→a
In words: The limit of a quotient is the quotient of the limits
x2 + 2
Example: lim
x→0 1 + cos x
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lim (x 2 + 2)
=
x→0
lim (1 + cos x)
x→0
Math 121 - Summary of Limit Laws
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Power Law
n
lim [f (x)] =
h
x→a
in
lim f (x) where n is a positive integer.
x→a
In words: The limit of a power is the power of the limit
Example: lim [x + tan x]1000 =
x→π
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h
i1000
lim (x + tan x)
x→π
Math 121 - Summary of Limit Laws
May 7 2012
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Root Law
lim
x→a
q
p
n
f (x) = n lim f (x) where n is a positive integer, and where
x→a
lim f (x) > 0 if n is even.
x→a
In words: The limit of a root is the root of the limit
Example: lim
x→1
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p
x2
+
5x 3
=
r
lim x 2 + 5x 3
x→1
Math 121 - Summary of Limit Laws
May 7 2012
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Particular Limit Results
G.Pugh (VIU)
Math 121 - Summary of Limit Laws
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Constants
lim c = c
x→a
√
Example: lim
x→3
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√
2π =
2π
Math 121 - Summary of Limit Laws
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Limit of f (x) = x
lim x = a
x→a
Example: lim x = 5
x→5
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Math 121 - Summary of Limit Laws
May 7 2012
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Polynomials
Using the Sum, Difference, Constant Multiplier and Power Laws:
If f (x) is a polynomial, (for eg. f (x) = 5x 3 − πx 2 − 12 ), then
lim f (x) = f (a) .
x→a
Example:
lim 5x 3 − πx 2 −
x→−1
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1
1
11
= 5(−1)3 − π(−1)2 − = − π −
2
2
2
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Rational Functions
Using the previous result and the Quotient Law:
If f (x) and g(x) are polynomials and g(a) 6= 0 then
f (x)
f (a)
lim
.
=
x→a g(x)
g(a)
2x 3 − x
2(2)3 − (2)
14
=
=
=2
3(2) + 1
7
x→2 3x + 1
Example: lim
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Trigonometric Functions
lim sin (x) = sin (a)
x→a
lim cos (x) = cos (a)
x→a
Example: lim sin (x) = sin (π/6) =
x→π/6
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1
2
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Direct Substitution Property
Putting together these limit results we have the Direct Substitution
Property :
If f (x) is a function defined using sums, differences, products or
quotients involving polynomials, sin (x), or cos (x), and
if a is in the domain of f (x) (that is, f (a) is defined)
then
lim f (x) = f (a)
x→a
Example:
−2x 3 − sin2 (x)
−2π 3 − sin2 (π)
−2π 3 − 0
lim
=
=
= 2π 3
x→π
cos3 (x)
cos3 (π)
(−1)3
G.Pugh (VIU)
Math 121 - Summary of Limit Laws
May 7 2012
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Some Advice
When evaluating limits, try to apply the Direct Substitution Property
first.
If direct substitution fails, then resort to more sophisticated techniques.
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Math 121 - Summary of Limit Laws
May 7 2012
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