Properties of Limits
Notation of Limit
The limit of a function is designated by f (x) → L as x → a or using the limit notation:
\[\lim\limits_{x \to a} f\left( x \right) = L.\]
Below we assume that the limits of functions \(\lim\limits_{x \to a} f\left( x \right),\) \(\lim\limits_{x \to a} g\left( x \right),\) \(\lim\limits_{x \to a} {f_1}\left( x \right),\) \(\ldots,\) \(\lim\limits_{x \to a} {f_n}\left( x \right)\) exist.
Sum Rule
This rule states that the limit of the sum of two functions is equal to the sum of their limits:
\[\lim\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \lim\limits_{x \to a} f\left( x \right) + \lim\limits_{x \to a} g\left( x \right).\]
Extended Sum Rule
\[\lim\limits_{x \to a} \left[ {{f_1}\left( x \right) + \ldots + {f_n}\left( x \right)} \right] = \lim\limits_{x \to a} {f_1}\left( x \right) + \ldots + \lim\limits_{x \to a} {f_n}\left( x \right).\]
Constant Function Rule
The limit of a constant function is the constant:
\[\lim\limits_{x \to a} C = C.\]
Constant Multiple Rule
The limit of a constant times a function is equal to the product of the constant and the limit of the function:
\[\lim\limits_{x \to a} kf\left( x \right) = k\lim\limits_{x \to a} f\left( x \right).\]
Product Rule
This rule says that the limit of the product of two functions is the product of their limits (if they exist):
\[\lim\limits_{x \to a} \left[ {f\left( x \right)g\left( x \right)} \right] = \lim\limits_{x \to a} f\left( x \right) \cdot \lim\limits_{x \to a} g\left( x \right).\]
Extended Product Rule
\[\lim\limits_{x \to a} \left[ {{f_1}\left( x \right){f_2}\left( x \right) \cdots {f_n}\left( x \right)} \right] = \lim\limits_{x \to a} {f_1}\left( x \right) \cdot \lim\limits_{x \to a} {f_2}\left( x \right) \cdots \lim\limits_{x \to a} {f_n}\left( x \right).\]
Quotient Rule
The limit of quotient of two functions is the quotient of their limits, provided that the limit in the denominator function is not zero:
\[\lim\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\lim\limits_{x \to a} f\left( x \right)}}{{\lim\limits_{x \to a} g\left( x \right)}},\;\;\; \text{if}\;\;\lim\limits_{x \to a} g\left( x \right) \ne 0.\]
Power Rule
\[\lim\limits_{x \to a} {\left[ {f\left( x \right)} \right]^p} = {\left[ {\lim\limits_{x \to a} f\left( x \right)} \right]^p},\]
where the power \(p\) can be any real number. In particular,
\[\lim\limits_{x \to a} \sqrt[p]{{f\left( x \right)}} = \sqrt[p]{{\lim\limits_{x \to a} f\left( x \right)}}.\]
If \(f\left( x \right) = x^n,\) then
\[\lim\limits_{x \to a} {x^n} = {a^n},\;n = 0, \pm 1, \pm 2, \ldots \;\;\; \text{and}\;\;a \ne 0,\;\;\text{if}\;\;n \le 0.\]
This is a special case of the previous property.
Limit of an Exponential Function
\[\lim\limits_{x \to a} {b^{f\left( x \right)}} = {b^{\lim\limits_{x \to a} f\left( x \right)}},\]
where the base \(b \gt 0.\)
Limit of a Logarithm of a Function
\[\lim\limits_{x \to a} \left[ {\log _b f\left( x \right)} \right] = \log_b \left[ {\lim\limits_{x \to a} f\left( x \right)} \right],\]
where the base \(b \gt 0.\)
The Squeeze Theorem
Suppose that \(g\left( x \right) \le f\left( x \right) \le h\left( x \right)\) for all \(x\) close to \(a,\) except perhaps for \(x = a.\) If
\[\lim\limits_{x \to a} g\left( x \right) = \lim\limits_{x \to a} h\left( x \right) = L,\]
then
\[\lim\limits_{x \to a} f\left( x \right) = L.\]
The idea here is that the function \(f\left( x \right)\) is squeezed between two other functions having the same limit \(L.\)
Solved Problems
Example 1.
Find the limit \[\lim\limits_{x \to 10} \left( {2x\lg {x^3}} \right).\]
Solution.
\[\lim\limits_{x \to 10} \left( {2x\lg {x^3}} \right) = \lim\limits_{x \to 10} 2x \cdot \lim\limits_{x \to 10} \lg {x^3} = 2\lim\limits_{x \to 10} x \cdot \lg \left( {\lim\limits_{x \to 10} {x^3}} \right) = 2 \cdot 10 \cdot \lg 1000 = 20 \cdot 3 = 60. \]
Example 2.
Find the limit \[\lim\limits_{x \to 9} {\frac{{4{x^2}}}{{1 + \sqrt x }}}.\]
Solution.
Using the properties of limits (the sum rule, the power rule, and the quotient rule), we get
\[\lim\limits_{x \to 9} \frac{{4{x^2}}}{{1 + \sqrt x }} = \frac{{\lim\limits_{x \to 9} 4{x^2}}}{{\lim\limits_{x \to 9} \left( {1 + \sqrt x } \right)}} = \frac{{4\lim\limits_{x \to 9} {x^2}}}{{\lim\limits_{x \to 9} 1 + \lim\limits_{x \to 9} \sqrt x }} = \frac{{4 \cdot {9^2}}}{{1 + \sqrt 9 }} = 81.\]
See more problems on Page 2.
