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.
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.
\[ \lim_{{x \to 5}} m(x) = \sqrt{{9}} = 3 \]
