Calculus

Limits and Continuity of Functions

Limits and Continuity Logo

Properties of Limits

Notation of Limit

The limit of a function is designated by \(f\left( x \right) \to L\) as \(x \to a\) 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

Click or tap a problem to see the solution.

Example 1

Find the limit \[\lim\limits_{x \to 10} \left( {2x\lg {x^3}} \right).\]

Example 2

Find the limit \[\lim\limits_{x \to 9} {\frac{{4{x^2}}}{{1 + \sqrt x }}}.\]

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.

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