Properties of limits are presented along with some examples and detailed solutions.
Theorem: If f and g are two functions and both lim_{x→a} f(x) and lim_{x→a} g(x) exist, then
**1. lim [ f(x) + g(x) ] = lim f(x) + lim g(x) : The limit of the sum of two functions is the sum of their limits.**
__Example 1:__ Calculate 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
**2. lim [ f(x) - g(x) ] = lim f(x) - lim g(x) : The limit of the difference of two functions is the difference of their limits.**
__Example 2:__ Calculate 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
**3. lim [ f(x) * g(x) ] = lim f(x) * lim g(x) : The limit of the product of two functions is the product of their limits.
**
__Example 3:__ Calculate 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
limx→- 5 x = - 5
Hence,
limx→-5 m(x) = 3*(- 5) = - 15
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**4. lim [ f(x) / g(x) ] = lim f(x) / lim g(x) ; if lim g(x) is not equal to zero. 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.**
__Example 4:__ Calculate 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
**5. lim nth root [ f(x) ] = nth root [ lim f(x) ]. If n is even, lim f(x) has to be positive. 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.**
__Example 5:__ Calculate lim_{x→5} m(x) where m(x) is given by
m(x) = SQRT[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) = SQRT(9) = 3
More on limits
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