Explanation Given information: The functions are given by f ( x ) = x 2 − 1 and g ( x ) = ln x at x = 1 . Calculation: Consider, f ( x ) = x 2 − 1 The point of tangency is f ( 1 ) = 1 2 − 1 = 0 → ( 1 , 0 ) Slope of tangent, f ′ ( x ) = 2 x Then, slope of tangent at x = 1 → f ′ ( 1 ) = 2 So, y = m x + c 0 = 2 × 1 + c → c = − 2 y = 2 x − 2 Similarly consider, g ( x ) = ln x The point of tangency is g ( 1 ) = ln ( 1 ) <

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ISBN: 9781429241861

Author: Laura Taalman, Peter Kohn

Publisher: W. H. Freeman

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Chapter 3.6, Problem 5C

To determine

To calculate: Calculate the equations of the tangent lines. Then using graphs prove that limx→1f(x)g(x) is equal to the limit of the quotient of tangent lines as x→1 .

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Chapter 3.6, Problem 4C

Chapter 3.6, Problem 6C

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Chapter 3 Solutions

Calculus

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Calculate the equations of the tangent lines. Then using graphs prove that lim x → 1 f ( x ) g ( x ) is equal to the limit of the quotient of tangent lines as x → 1 .

Solution Summary: The author calculates the equations of the tangent lines using graphs. They prove that undersetxto 1mathrmlim

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To calculate: Calculate the equations of the tangent lines. Then using graphs prove that limx→1f(x)g(x) is equal to the limit of the quotient of tangent lines as x→1 .

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Chapter 3 Solutions