Properties of Limits of Functions in Calculus

Properties of limits of functions, in the form of theorems, are presented along with some examples of applications and detailed solutions.

Theorem: If \( f \) and \( g \) are two functions and both \( \lim_{{x \to a}} f(x) \) and \( \lim_{{x \to a}} g(x) \) exist, then

Property 1: The limit of the sum of two functions is the sum of their limits.

\[ \lim_{{x \to a}} [ f(x) + g(x) ] = \lim_{{x \to a}} f(x) + \lim_{{x \to a}} g(x) \]

Example 1

Calculate \( \lim_{{x \to -2}} h(x) \) where \( h(x) \) is given by \[ h(x) = x + 5 \] Solution to Example 1:
We may consider \( h(x) \) as the sum of \( f(x) = x \) and \( g(x) = 5 \) and apply theorem 1 above \[ \lim_{{x \to -2}} h(x) = \lim_{{x \to -2}} x + \lim_{{x \to -2}} 5 \] \( x \) and \( 5 \) are basic functions and their limits are known. \[ \lim_{{x \to -2}} x = -2 \] and \[ \lim_{{x \to -2}} 5 = 5 \] Hence, \[ \lim_{{x \to -2}} h(x) = -2 + 5 = 3 \]

Property 2: The limit of the difference of two functions is the difference of their limits.

\[ \lim_{{x \to a}} [ f(x) - g(x) ] = \lim_{{x \to a}} f(x) - \lim_{{x \to a}} g(x) \] :

Example 2

Calculate \( \lim_{{x \to 10}} h(x) \) where \( h(x) \) is given by
    \[ h(x) = x - 7 \] Solution to Example 2:
    We may consider \( h(x) \) as the difference of \( f(x) = x \) and \( g(x) = 7 \) and apply theorem 2 above \[ \lim_{{x \to 10}} h(x) = \lim_{{x \to 10}} x - \lim_{{x \to 10}} 7 \] \( x \) and \( 7 \) are basic functions with known limits. \[ \lim_{{x \to 10}} x = 10 \] and \[ \lim_{{x \to 10}} 7 = 7 \] Hence, \[ \lim_{{x \to 10}} h(x) = 10 - 7 = 3 \]

Property 3: The limit of the product of two functions is the product of their limits

\[ \lim_{{x \to a}} [ f(x) \times g(x) ] = \lim_{{x \to a}} f(x) \times \lim_{{x \to a}} g(x) \] :

    Example 3

    Calculate \( \lim_{{x \to -5}} m(x) \) where \( m(x) \) is given by \[ m(x) = 3x \] Solution to Example 3:
    Let \( m(x) = f(x) \times g(x) \), where \( f(x) = 3 \) and \( g(x) = x \) and apply theorem 3 above \[ \lim_{{x \to -5}} m(x) = \lim_{{x \to -5}} 3 \times \lim_{{x \to -5}} x \] 3 is a constant function and \( x \) is also a basic function with known limits. \[ \lim_{{x \to -5}} 3 = 3 \] and \[ \lim_{{x \to -5}} x = -5 \] Hence, \[ \lim_{{x \to -5}} m(x) = 3 \times (-5) = -15 \]

Property 4: The limit of the quotient of two functions is the quotient of their limits if the limit in the denominator is not equal to 0.

\[ \lim_{{x \to a}} \frac{{f(x)}}{{g(x)}} = \frac{{\lim_{{x \to a}} f(x)}}{{\lim_{{x \to a}} g(x)}} \] ; if \( \lim_{{x \to a}} g(x) \) is not equal to zero.

    Example 4

    Calculate \( \lim_{{x \to 3}} r(x) \) where \( r(x) \) is given by \[ r(x) = \frac{{3 - x}}{x} \] Solution to Example 4:
    Let \( r(x) = \frac{{f(x)}}{{g(x)}} \), where \( f(x) = 3 - x \) and \( g(x) = x \) and apply theorem 4 above \[ \lim_{{x \to 3}} r(x) = \frac{{\lim_{{x \to 3}} (3 - x)}}{{\lim_{{x \to 3}} x}} \] \( 3 - x \) is the difference of two basic functions and \( x \) is also a basic function. \[ \lim_{{x \to 3}} (3 - x) = 3 - 3 = 0 \] and \[ \lim_{{x \to 3}} x = 3 \] Hence, \[ \lim_{{x \to 3}} r(x) = \frac{{0}}{{3}} = 0 \]

Property 5: The limit of the nth root of a function is the nth root of the limit of the function, if the nth root of the limit is a real number.

\[ \lim_{{x \to a}} \sqrt[n]{{f(x)}} = \sqrt[n]{{\lim_{{x \to a}} f(x)}} \]. If \( n \) is even, \( \lim_{{x \to a}} f(x) \) has to be positive.

    Example 5

    Calculate \( \lim_{{x \to 5}} m(x) \) where \( m(x) \) is given by \[ m(x) = \sqrt{{2x - 1}} \] Solution to Example 5:
    Let \( f(x) = 2x - 1 \) and find its limit applying the difference and product theorems above \[ \lim_{{x \to 5}} f(x) = 2 \times 5 - 1 = 9 \] We now apply theorem 5 since the square root of 9 is a real number. \[ \lim_{{x \to 5}} m(x) = \sqrt{{9}} = 3 \]

More Links on limits

Calculus Tutorials and Problems
Limits of Absolute Value Functions Questions

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