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 ## Property 1: The limit of the sum of two functions is the sum of their limits.lim [ f(x) + g(x) ] = lim f(x) + lim g(x)## Example 1Calculate lim_{x→-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→-2} h(x) = lim_{x→-2} x + lim_{x→-2} 5
x and 5 are basic functions and their limits are known. lim _{x→-2} x = -2
and lim _{x→-2} 5 = 5
Hence, lim _{x→-2} h(x) = -2 + 5 = 3
## Property 2: The limit of the difference of two functions is the difference of their limits.lim [ f(x) - g(x) ] = lim f(x) - lim g(x) :## Example 2Calculate lim_{x→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→10} h(x) = lim_{x→10} x - lim_{x→10} 7
x and 7 are basic functions with known limits. lim _{x→10} x = 10
and lim _{x→10} 7 = 7
Hence, lim _{x→10} h(x) = 10 - 7 = 3
## Property 3: The limit of the product of two functions is the product of their limitslim [ f(x) × g(x) ] = lim f(x) × lim g(x) :## Example 3Calculate lim_{x→-5} m(x) where m(x) is given by
m(x) = 3 x Solution to Example 3:
Let m(x) = f(x) * g(x), where f(x) = 3 and g(x) = x and apply theorem 3 above lim _{x→-5} m(x) = lim_{x→-5} 3 * lim_{x→-5} x
3 is a constant function and x is also a basic function with known limits. lim _{x→}3 = 3
and lim _{x→- 5} x = - 5
Hence, lim _{x→-5} m(x) = 3*(- 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 [ f(x) / g(x) ] = lim f(x) / lim g(x) ; if lim g(x) is not equal to zero.## Example 4Calculate lim_{x→3} r(x) where r(x) is given by
r(x) = (3 - x) / x Solution to Example 4:
Let r(x) = f(x) / g(x), where f(x) = 3 - x and g(x) = x and apply theorem 4 above lim _{x→3} r(x) = lim_{x→3} (3 - x) / lim_{x→3} x
3 - x is the difference of two basic functions and x is also a basic function. lim _{x→3} (3 - x) = 3 - 3 = 0
and lim _{x→3} x = 3
Hence, lim _{x→3} r(x) = 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^{n}√[ f(x) ] = ^{n}√[ lim f(x) ]. If n is even, lim f(x) has to be positive.
## Example 5Calculate lim_{x→5} m(x) where m(x) is given by
m(x) = √[2 x - 1] Solution to Example 5:
Let f(x) = 2 x - 1 and find its limit applying the difference and product theorems above lim _{x→5} f(x) = 2*5 - 1 = 9
We now apply theorem 5 since the square root of 9 is a real number. lim _{x→5} m(x) = √(9) = 3
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