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Key concepts

  • The limit laws allow us to evaluate limits of functions without having to go through step-by-step processes each time.
  • For polynomials and rational functions, lim x a f ( x ) = f ( a ) .
  • You can evaluate the limit of a function by factoring and canceling, by multiplying by a conjugate, or by simplifying a complex fraction.
  • The squeeze theorem allows you to find the limit of a function if the function is always greater than one function and less than another function with limits that are known.

Key equations

  • Basic Limit Results
    lim x a x = a lim x a c = c
  • Important Limits
    lim θ 0 sin θ = 0
    lim θ 0 cos θ = 1
    lim θ 0 sin θ θ = 1
    lim θ 0 1 cos θ θ = 0

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

lim x 0 ( 4 x 2 2 x + 3 )

Use constant multiple law and difference law: lim x 0 ( 4 x 2 2 x + 3 ) = 4 lim x 0 x 2 2 lim x 0 x + lim x 0 3 = 3

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lim x 1 x 3 + 3 x 2 + 5 4 7 x

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lim x −2 x 2 6 x + 3

Use root law: lim x −2 x 2 6 x + 3 = lim x −2 ( x 2 6 x + 3 ) = 19

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lim x −1 ( 9 x + 1 ) 2

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In the following exercises, use direct substitution to evaluate each limit.

lim x −2 ( 4 x 2 1 )

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lim x 0 1 1 + sin x

1

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lim x 2 e 2 x x 2

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lim x 1 2 7 x x + 6

5 7

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In the following exercises, use direct substitution to show that each limit leads to the indeterminate form 0 / 0 . Then, evaluate the limit.

lim x 4 x 2 16 x 4

lim x 4 x 2 16 x 4 = 16 16 4 4 = 0 0 ; then, lim x 4 x 2 16 x 4 = lim x 4 ( x + 4 ) ( x 4 ) x 4 = 8

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lim x 2 x 2 x 2 2 x

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lim x 6 3 x 18 2 x 12

lim x 6 3 x 18 2 x 12 = 18 18 12 12 = 0 0 ; then, lim x 6 3 x 18 2 x 12 = lim x 6 3 ( x 6 ) 2 ( x 6 ) = 3 2

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lim x 0 ( 1 + h ) 2 1 h

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lim x 9 t 9 t 3

lim x 9 t 9 t 3 = 9 9 3 3 = 0 0 ; then, lim t 9 t 9 t 3 = lim t 9 t 9 t 3 t + 3 t + 3 = lim t 9 ( t + 3 ) = 6

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lim h 0 1 a + h 1 a h , where a is a real-valued constant

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lim θ π sin θ tan θ

lim θ π sin θ tan θ = sin π tan π = 0 0 ; then, lim θ π sin θ tan θ = lim θ π sin θ sin θ cos θ = lim θ π cos θ = −1

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lim x 1 x 3 1 x 2 1

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lim x 1 / 2 2 x 2 + 3 x 2 2 x 1

lim x 1 / 2 2 x 2 + 3 x 2 2 x 1 = 1 2 + 3 2 2 1 1 = 0 0 ; then, lim x 1 / 2 2 x 2 + 3 x 2 2 x 1 = lim x 1 / 2 ( 2 x 1 ) ( x + 2 ) 2 x 1 = 5 2

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lim x −3 x + 4 1 x + 3

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In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of [link] to simplify the function to help determine the limit.

lim x −2 2 x 2 + 7 x 4 x 2 + x 2

−∞

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lim x −2 + 2 x 2 + 7 x 4 x 2 + x 2

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lim x 1 2 x 2 + 7 x 4 x 2 + x 2

−∞

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lim x 1 + 2 x 2 + 7 x 4 x 2 + x 2

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In the following exercises, assume that lim x 6 f ( x ) = 4 , lim x 6 g ( x ) = 9 , and lim x 6 h ( x ) = 6 . Use these three facts and the limit laws to evaluate each limit.

lim x 6 2 f ( x ) g ( x )

lim x 6 2 f ( x ) g ( x ) = 2 lim x 6 f ( x ) lim x 6 g ( x ) = 72

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lim x 6 g ( x ) 1 f ( x )

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lim x 6 ( f ( x ) + 1 3 g ( x ) )

lim x 6 ( f ( x ) + 1 3 g ( x ) ) = lim x 6 f ( x ) + 1 3 lim x 6 g ( x ) = 7

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lim x 6 ( h ( x ) ) 3 2

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lim x 6 g ( x ) f ( x )

lim x 6 g ( x ) f ( x ) = lim x 6 g ( x ) lim x 6 f ( x ) = 5

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lim x 6 x · h ( x )

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lim x 6 [ ( x + 1 ) · f ( x ) ]

lim x 6 [ ( x + 1 ) f ( x ) ] = ( lim x 6 ( x + 1 ) ) ( lim x 6 f ( x ) ) = 28

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lim x 6 ( f ( x ) · g ( x ) h ( x ) )

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[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

f ( x ) = { x 2 , x 3 x + 4 , x > 3

  1. lim x 3 f ( x )
  2. lim x 3 + f ( x )


The graph of a piecewise function with two segments. The first is the parabola x^2, which exists for x<=3. The vertex is at the origin, it opens upward, and there is a closed circle at the endpoint (3,9). The second segment is the line x+4, which is a linear function existing for x > 3. There is an open circle at (3, 7), and the slope is 1.
a. 9; b. 7

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g ( x ) = { x 3 1 , x 0 1 , x > 0

  1. lim x 0 g ( x )
  2. lim x 0 + g ( x )
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h ( x ) = { x 2 2 x + 1 , x < 2 3 x , x 2

  1. lim x 2 h ( x )
  2. lim x 2 + h ( x )


The graph of a piecewise function with two segments. The first segment is the parabola x^2 – 2x + 1, for x < 2. It opens upward and has a vertex at (1,0). The second segment is the line 3-x for x>= 2. It has a slope of -1 and an x intercept at (3,0).
a. 1; b. 1

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In the following exercises, use the following graphs and the limit laws to evaluate each limit.

Two graphs of piecewise functions. The upper is f(x), which has two linear segments. The first is a line with negative slope existing for x < -3. It goes toward the point (-3,0) at x= -3. The next has increasing slope and goes to the point (-3,-2) at x=-3. It exists for x > -3. Other key points are (0, 1), (-5,2), (1,2), (-7, 4), and (-9,6). The lower piecewise function has a linear segment and a curved segment. The linear segment exists for x < -3 and has decreasing slope. It goes to (-3,-2) at x=-3. The curved segment appears to be the right half of a downward opening parabola. It goes to the vertex point (-3,2) at x=-3. It crosses the y axis a little below y=-2. Other key points are (0, -7/3), (-5,0), (1,-5), (-7, 2), and (-9, 4).

lim x −3 + ( f ( x ) + g ( x ) )

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lim x −3 ( f ( x ) 3 g ( x ) )

lim x −3 ( f ( x ) 3 g ( x ) ) = lim x −3 f ( x ) 3 lim x −3 g ( x ) = 0 + 6 = 6

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lim x 0 f ( x ) g ( x ) 3

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lim x −5 2 + g ( x ) f ( x )

lim x −5 2 + g ( x ) f ( x ) = 2 + ( lim x −5 g ( x ) ) lim x −5 f ( x ) = 2 + 0 2 = 1

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lim x 1 ( f ( x ) ) 2

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lim x 1 f ( x ) g ( x )

lim x 1 f ( x ) g ( x ) 3 = lim x 1 f ( x ) lim x 1 g ( x ) 3 = 2 + 5 3 = 7 3

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lim x −7 ( x · g ( x ) )

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lim x −9 [ x · f ( x ) + 2 · g ( x ) ]

lim x −9 ( x f ( x ) + 2 g ( x ) ) = ( lim x −9 x ) ( lim x −9 f ( x ) ) + 2 lim x −9 ( g ( x ) ) = ( −9 ) ( 6 ) + 2 ( 4 ) = −46

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For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions f ( x ) , g ( x ) , and h ( x ) when possible.

[T] True or False? If 2 x 1 g ( x ) x 2 2 x + 3 , then lim x 2 g ( x ) = 0 .

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[T] lim θ 0 θ 2 cos ( 1 θ )

The limit is zero.
The graph of three functions over the domain [-1,1], colored red, green, and blue as follows: red: theta^2, green: theta^2 * cos (1/theta), and blue: - (theta^2). The red and blue functions open upwards and downwards respectively as parabolas with vertices at the origin. The green function is trapped between the two.

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lim x 0 f ( x ) , where f ( x ) = { 0 , x rational x 2 , x irrrational

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[T] In physics, the magnitude of an electric field generated by a point charge at a distance r in vacuum is governed by Coulomb’s law: E ( r ) = q 4 π ε 0 r 2 , where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and 1 4 π ε 0 is Coulomb’s constant: 8.988 × 10 9 N · m 2 / C 2 .

  1. Use a graphing calculator to graph E ( r ) given that the charge of the particle is q = 10 −10 .
  2. Evaluate lim r 0 + E ( r ) . What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

a.
A graph of a function with two curves. The first is in quadrant two and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to negative infinity. The second is in quadrant one and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to infinity.
b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

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[T] The density of an object is given by its mass divided by its volume: ρ = m / V .

  1. Use a calculator to plot the volume as a function of density ( V = m / ρ ) , assuming you are examining something of mass 8 kg ( m = 8 ).
  2. Evaluate lim x 0 + V ( r ) and explain the physical meaning.
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Practice Key Terms 9

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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